The Unapologetic Mathematician

Some Continuous Duals

I really wish I could just say $L^p$ in post titles.

Anyway, I want to investigate the continuous dual of $L^p(\mu)$ for $1\leq p<\infty$. That is, we’re excluding the case where either $p$ (but not its Hölder conjugate $q$) is infinite. And I say that when $(X,\mathcal{S},\mu)$ is $\sigma$-finite, the space $L^p(\mu)'$ of bounded linear functionals on $L^p(\mu)$ is isomorphic to $L^q(\mu)$.

First, I’m going to define a linear map $\kappa_p:L^q\to\left(L^p\right)'$. Given a function $g\in L^q$, let $\kappa_p(g)$ be the linear functional defined for any $f\in L^p$ by

$\displaystyle\left[\kappa_p(g)\right](f)=\int fg\,d\mu$

It’s clear from the linearity of multiplication and of the integral itself, that this is a linear functional on $L^p$. Hölder’s inequality itself shows us that not only does the integral on the right exist, but

$\displaystyle\lvert\left[\kappa_p(g)\right](f)\rvert\leq\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$

That is, $\kappa_p(g)$ is a bounded linear functional, and the operator norm $\lVert\kappa_p(g)\rVert_\text{op}$ is at most the $L^q$ norm of $g$. The extremal case of Hölder’s inequality shows that there is some $f$ for which this is an equality, and thus we conclude that $\lVert\kappa_p(g)\rVert_\text{op}=\lVert g\rVert_q$. That is, $\kappa_p:L^q\to\left(L^p\right)'$ is an isometry of normed vector spaces. Such a mapping has to be an injection, because if $\kappa_p(g)=0$ then $0=\lVert\kappa_p(g)\rVert_\text{op}=\lVert g\rVert_q$, which implies that $g=0$.

Now I say that $\kappa_p$ is also a surjection. That is, any bounded linear functional $\Lambda:L^p\to\mathbb{R}$ is of the form $\kappa_p(g)$ for some $g\in L^q$. Indeed, if $\Lambda=0$ then we can just pick $g=0$ as a preimage. Thus we may assume that $\Lambda$ is a nonzero bounded linear functional on $L^p$, and $\lVert\Lambda\rVert_\text{op}>0$. We first deal with the case of a totally finite measure space.

In this case, we define a set function on measurable sets by $\lambda(E)=\Lambda(\chi_E)$. It’s straightforward to see that $\lambda$ is additive. To prove countable additivity, suppose that $E$ is the countable disjoint union of a sequence $\{E_n\}$. If we write $A_k$ for the union of $E_1$ through $E_k$, we find that

$\displaystyle\lVert\chi_E-\chi_{A_k}\rVert_p=\left(\mu(E\setminus A_k)\right)^\frac{1}{p}\to0$

Since $\Lambda$ is continuous, we conclude that $\lambda(A_k)\to\lambda(E)$, and thus that $\lambda$ is a (signed) measure. It should also be clear that $\mu(E)=0$ implies $\lambda(E)=0$, and so $\lambda\ll\mu$. The Radon-Nikodym theorem now tells us that there exists an integrable function $g$ so that

$\displaystyle\Lambda(\chi_E)=\lambda(E)=\int\limits_Eg\,d\mu=\int\chi_Eg\,d\mu$

Linearity tells us that

$\displaystyle\Lambda(f)=\int fg\,d\mu$

for simple functions $f$, and also for every $f\in L^\infty(\mu)$, since each such function is the uniform limit of simple functions. We want to show that $g\in L^q$.

If $p=1$, then we must show that $g$ is essentially bounded. In this case, we find

$\displaystyle\left\lvert\int\limits_Eg\,d\mu\right\rvert\leq\lVert\Lambda\rVert_\text{op}\,\lVert\chi_E\rVert_1=\lVert\Lambda\rVert_\text{op}\mu(E)$

for every measurable $E$, from which we conclude that $\lvert g(x)\rvert\leq\lVert\Lambda\rVert_\text{op}$ a.e., or else we could find some set on which this inequality was violated. Thus $\lVert g\rVert_\infty\leq\lVert\Lambda\rVert_\text{op}$.

For other $p$, we can find a measurable $\alpha$ with $\lvert\alpha\rvert=1$ so that $\alpha g=\lvert g\rvert$. Setting $E_n=\{x\in X\vert n\geq\lvert g(x)\rvert\}$ and defining $f=\chi_{E_n}\lvert g\rvert^{q-1}\alpha$, we find that $\lvert f\rvert^p=\lvert g\rvert^q$ on $E_n$, $f\in L^\infty$, and so

$\displaystyle\int\limits_{E_n}\lvert g\rvert^q\,d\mu=\int\limits_Xfg\,d\mu=\Lambda(f)\leq\lVert\Lambda\rVert_\text{op}=\lVert\Lambda\rVert_\text{op}\left(\int\limits_{E_n}\lvert g\rvert^q\,d\mu\right)^\frac{1}{p}$

We thus find

$\displaystyle\left(\int\limits_{E_n}\lvert g\rvert^q\,d\mu\right)^\frac{1}{q}=\left(\int\limits_{E_n}\lvert g\rvert^q\,d\mu\right)^{1-\frac{1}{p}}\leq\lVert\Lambda\rVert_\text{op}$

and thus

$\displaystyle\int\limits_X\chi_{E_n}\lvert g\rvert^q\,d\mu\leq\lVert\Lambda\rVert_\text{op}^q$

Applying the monotone convergence theorem as $n\to\infty$ we find that $\lVert g\rVert_q\leq\lVert\Lambda\rVert_\text{op}$.

Thus in either case we’ve found a $g\in L^q$ so that $\Lambda=\kappa_p(g)$.

In the $\sigma$-finite case, we can write $X$ as the countable disjoint union of sets $X_i$ with $\mu(X_i)<\infty$. We let $Y_k$ be the union of the first $k$ of these sets. We note that $\lVert\chi_E f\rVert_p\leq\lVert f\rVert_p$ for every measurable set $E$, so $f\mapsto\Lambda(\chi_Ef)$ is a linear functional on $L^p$ of norm at most $\lVert\Lambda\rVert_\text{op}$. The finite case above shows us that there are functions $g_i$ on $X_i$ so that

$\displaystyle\Lambda(\chi_{X_i}f)=\int\limits_{X_i}fg_i\,d\mu$.

We can define $g_i(x)=0$ if $x\notin X_i$, and let $g$ be the sum of all these $g_i$. We see that

$\displaystyle\Lambda(\chi_{Y_k}f)=\int\limits_{Y_k}f(g_1+\dots+g_k)\,d\mu$

for every $f\in L^p$, and since $\mu(Y_k)<\infty$ we find that $\lVert g_1+\dots+g_k\rVert_q\leq\lVert\Lambda\rVert_\text{op}$. Then Fatou’s lemma shows us that $\lVert g\rVert_q\leq\lVert\Lambda\rVert_\text{op}$. Thus the $\sigma$-finite case is true as well.

One case in particular is especially worthy of note: since $2$ is Hölder-coonjugate to itself, we find that $L^2$ is isomorphic to its own continuous dual space in the same way that a finite-dimensional inner-product space is isomorphic to its own dual space.

September 3, 2010

Bounded Linear Transformations

In the context of normed vector spaces we have a topology on our spaces and so it makes sense to ask that maps $f:V\to W$ between them be continuous. In the finite-dimensional case, all linear functions are continuous, so this hasn’t really come up before in our study of linear algebra. But for functional analysis, it becomes much more important.

Now, really we only need to require continuity at one point — the origin, to be specific — because if it’s continuous there then it’ll be continuous everywhere. Indeed, continuity at $v'$ means that for any $\epsilon>0$ there is a $\delta>0$ so that $\lVert v-v'\rVert_V<\delta$ implies $\lVert f(v)-f(v')\rVert_W=\lVert f(v-v')\rVert_W<\epsilon$. In particular, if $v'=0$, then this means $\lVert v\rVert_V<\delta$ implies $\lVert f(v)\rVert_W<\epsilon$. Clearly if this holds, then the general version also holds.

But it turns out that there’s another equivalent condition. We say that a linear transformation $f:V\to W$ is “bounded” if there is some $M>0$ such that $\lVert f(v)\rVert_W\leq M\lVert v\rVert_V$ for all $v\in V$. That is, the factor by which $f$ stretches the length of a vector is bounded. By linearity, we only really need to check this on the unit sphere $\lVert v\rVert_V=1$, but it’s often just as easy to test it everywhere.

Anyway, I say that a linear transformation is continuous if and only if it’s bounded. Indeed, if $f:V\to W$ is bounded, then we find

$\displaystyle M\lVert h \rVert_V\geq\lVert f(h)\rVert_W=\lVert f(v+h)-f(v)\rVert_W$

so as we let $h$ approach $0$ — as $v+h$ approaches $v$ — the difference between $f(v+h)$ and $f(v)$ approaches zero as well. And so $f$ is continuous.

Conversely, if $f$ is continuous, then it is bounded. Since it’s continuous, we let $\epsilon=1$ and find a $\delta$ so that $\lVert f(h)\rVert_W<1$ for all vectors $h$ with $\lVert h\rVert_V<\delta$. Thus for all nonzero $v\in V$ we find

\displaystyle\begin{aligned}\lVert f(v)\rVert_W&=\left\lVert\frac{\lVert v\rVert_V}{\delta}f\left(\delta\frac{v}{\lVert v\rvert_V}\right)\right\rVert_W\\&=\frac{\lVert v\rVert_V}{\delta}\left\lVert f\left(\delta\frac{v}{\lVert v\rVert_V}\right)\right\rVert_W\\&\leq\frac{\lVert v\rVert_V}{\delta}\,1\\&=\frac{1}{\delta}\lVert v\rVert_V\end{aligned}

Thus we can use $M=\frac{1}{\delta}$ and conclude that $f$ is bounded.

The least such $M$ that works in the condition for $f$ to be bounded is called the “operator norm” of $f$, which we write as $\lVert f\rVert_\text{op}$. It’s straightforward to verify that $\lVert cf\rVert_\text{op}=\lvert c\rvert\lVert f\rVert_\text{op}$, and that $\lVert f\rVert_\text{op}=0$ if and only if $f$ is the zero operator. It remains to verify the triangle identity.

Let’s say that we have bounded linear transformations $f:V\to W$ and $g:T\to W$ with operator norms $M=\lVert f\rVert_\text{op}$ and $N=\lVert g\rVert_\text{op}$, respectively. We will show that $M+N$ works as a bound for $f+g$, and thus conclude that $\lVert f+g\rVert_\text{op}\leq\lVert f\rVert_\text{op}+\lVert g\rVert_\text{op}$. Indeed, we check that

\displaystyle\begin{aligned}\lVert[f+g](v)\rVert_W&=\lVert f(v)+g(v)\rVert_W\\&\leq\lVert f(v)\rVert_W+\lVert g(v)\rVert_W\\&\leq M\lVert v\rVert_V+N\lVert v\rVert_V\\&=(M+N)\lVert v\rVert_V\end{aligned}

and our assertion follows. In particular, when our base field is itself a normed linear space (like $\mathbb{C}$ or $\mathbb{R}$ itself) we can conclude that the “continuous dual space$V'$ consisting of bounded linear functionals $\Lambda:V\to\mathbb{F}$ is a normed linear space using the operator norm on $V'$.

September 2, 2010

The Extremal Case of Hölder’s Inequality

We will soon need to know that Hölder’s inequality is in a sense the best we can do, at least for finite $p$. That is, not only do we know that for any $f\in L^p$ and $g\in L^q$ we have $\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$, but for any $f\in L^p$ there is some $g\in L^q$ for which we actually have equality. We will actually prove that

$\displaystyle\lVert f\rVert_p=\max\left\{\left\lvert\int fg\,d\mu\right\rvert\Big\vert\lVert g\rVert_q\leq1\right\}$

That is, not only is the integral bounded above by $\lVert f\rVert_p\lVert g\rVert_q$ — and thus by $\lVert f\rVert_p$ — but there actually exists some $g$ in the unit ball which achieves this maximum.

Hölder’s inequality tells us that

$\displaystyle\left\lvert\int fg\,d\mu\right\rvert\leq\int\lvert fg\rvert\,d\mu\leq\lVert f\rVert_p\lVert g\rVert_q\leq\lVert f\rVert_p$

so $\lVert f\rVert_p$ must be at least as big as every element of the given set. If $\lVert f\rVert_p=0$, then it’s clear that the asserted equality holds, since $f=0$ a.e., and so $0$ is the only element of the set on the right. Thus from here we can assume $\lVert f\rVert_p>0$.

We now define a function $g$. At every point $x$ where $f(x)=0$ we set $g(x)=0$ as well. At all other $x$ we define

$\displaystyle g(x)=\lVert f\rVert_p^{1-p}\frac{\lvert f(x)\rvert^p}{f(x)}$

In the case where $p=1$ we will verify that $\lVert g\rVert_\infty=1$. That is, the essential supremum of $g$ is $1$. And, indeed, we find that $g(x)=1$ at points where $f(x)>0$, and $g(x)=-1$ at points where $f(x)<0$.

If $1, then we check

\displaystyle\begin{aligned}\lVert g\rVert_q&=\left(\int\lvert g\rvert^q\,d\mu\right)^\frac{1}{q}\\&=\left(\int\lVert f\rVert_p^{-p}\lvert f\rvert^p\,d\mu\right)^\frac{1}{q}\\&=\left(\lVert f\rVert_p^{-p}\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{q}\\&=\left(\lVert f\rVert_p^{-p}\lVert f\rVert_p^p\right)^\frac{1}{q}\\&=1\end{aligned}

In either case, it’s easy to see that

$\displaystyle\int fg\,d\mu=\lVert f\rVert_p$

as asserted.

September 1, 2010 Posted by | Analysis, Measure Theory | 2 Comments

Some Banach Spaces

To complete what we were saying about the $L^p$ spaces, we need to show that they’re complete. As it turns out, we can adapt the proof that mean convergence is complete, but we will take a somewhat different approach. It suffices to show that for any sequence of functions $\{u_n\}$ in $L^p$ so that the series of $L^p$-norms converges

$\displaystyle\sum\limits_{n=1}^\infty\lVert u_n\rVert_p<\infty$

the series of functions converges to some function $f\in L^p$.

For finite $p$, Minkowski’s inequality allows us to conclude that

$\displaystyle\int\left(\sum\limits_{n=1}^\infty\lvert u_n\rvert\right)^p\,d\mu=\left(\sum\limits_{n=1}^\infty\lVert u_n\rVert_p\right)^p<\infty$

The monotone convergence theorem now tells us that the limiting function

$\displaystyle f=\sum\limits_{n=1}^\infty u_n$

is defined a.e., and that $f\in L^p$. The dominated convergence theorem can now verify that the partial sums of the series are $L^p$-convergent to $f$:

$\displaystyle\int\left\lvert f-\sum\limits_{k=1}^n u_k\right\rvert^p\,d\mu\leq\int\left(\sum\limits_{l=n+1}^\infty\lvert u_l\rvert\right)^p\to0$

In the case $p=\infty$, we can write $c_n=\lVert u_n\rVert_\infty$. Then $\lvert u_n(x)\rvert except on some set $E_n$ of measure zero. The union of all the $E_n$ must also be negligible, and so we can throw it all out and just have $\lvert u_n(x)\rvert. Now the series of the $c_n$ converges by assumption, and thus the series of the $u_n$ must converge to some function bounded by the sum of the $c_n$ (except on the union of the $E_n$).

August 31, 2010

The Supremum Metric

We can actually extend what we’ve been doing with Hölder’s inequality and Minkowski’s inequality a little further. Given a metric space $(X,\mathcal{S},\mu)$, we’ve already discussed the idea of an “essentially bounded” function — one for which there is some real constant $c$ so that $f(x)\leq c$ for almost all $x\in X$. We will write $L^\infty(X)$ for the collection of essentially bounded functions on the measure space. It should be clear that these form a vector space.

We also discussed the “essential supremum” $\text{ess sup}(\lvert f\rvert)$ of an essentially bounded function. We’ll now write this as $\lVert f\rVert_\infty$, suggesting that it’s a norm. And it’s clear that $\lVert cf\rVert_\infty=\lvert c\rvert\lVert f\rVert_\infty$, and that $\lVert f\rVert_\infty=0$ if and only if $f=0$ almost everywhere. Verifying the triangle identity is exactly Minkowski’s inequality.

And, indeed, we know that $\lvert f(x)\rvert\leq\lVert f\rVert_\infty$ and $\lvert g(x)\rvert\leq\lVert g\rVert_\infty$ a.e., so $\lvert f(x)+g(x)\rvert\leq\lvert f(x)\rvert+\lvert g(x)\rvert\leq\lVert f\rVert_\infty+\lVert g\rVert_\infty$ a.e., so whatever the least such essential upper bound is smaller still. That is, $\lVert f+g\rVert_\infty=\lVert f\rVert_\infty+\lVert g\rVert_\infty$.

Now for Hölder’s inequality. For this purpose we consider $\frac{1}{\infty}=0$, and thus $\frac{1}{1}+\frac{1}{\infty}=1$, which means that $1$ and $\infty$ are Hölder-conjugates. Thus our assertion is that if $f$ is integrable and $g$ is essentially bounded, then $fg$ is integrable and $\lVert fg\rVert_1\leq\lVert f\rVert_1\lVert g\rVert_\infty$. Indeed, we know that $\lvert g(x)\rvert\leq\lVert g\rVert_\infty$, and so $\lvert f(x)g(x)\rvert=\lvert f(x)\rvert\lvert g(x)\rvert\leq \lvert f(x)\rvert\lVert g\rVert_\infty$ — both inequalities holding almost everywhere. From this, we conclude that

\displaystyle\begin{aligned}\lVert fg\rVert_1&=\int\lvert fg\rvert\,d\mu\\&\leq\int\lvert f\rvert\lVert g\rVert_\infty\,d\mu\\&=\int\lvert f\rvert\,d\mu\lVert g\rVert_\infty\\&=\lVert f\rVert_1\lVert g\rVert_\infty\end{aligned}

as we asserted. From now on, we’ll allow $p=\infty$ (and $q=1$) whenever we’re talking about a Hölder-conjugate pair or $L^p$-space.

August 30, 2010 Posted by | Analysis, Measure Theory | 1 Comment

Minkowski’s Inequality

We continue our project to show that the $L^p$ spaces are actually Banach spaces with Minkowski’s inequality. This will allow us to conclude that $L^p$ is a normed vector space. It states that if $f$ and $g$ are both in $L^p$, then their sum $f+g$ is in $L^p$, and we have the inequality

$\displaystyle\lVert f+g\rVert_p\leq\lVert f\rVert_p+\lVert g\rVert_p$

We start by considering Hölder’s inequality in a toy space I’ll whip up right now. Take two isolated points, and let each one have measure $1$; the whole space of both points has measure $2$. A function is just an assignment of a pair of real values $(a_1,a_2)$, and integration just means adding them together. Hölder’s inequality for this space tells us that

$\displaystyle\lvert a_1b_1+a_2b_2\rvert\leq\left(\lvert a_1\rvert^p+\lvert a_2\rvert^p\right)^\frac{1}{p}\left(\lvert b_1\rvert^q+\lvert b_2\rvert^q\right)^\frac{1}{q}$

where $p$ and $q$ are Hölder-conjugate to each other. We can set $a_1=\lvert f\rvert^p$, $a_2=\lvert g\rvert^p$, and $b_1=b_2=\lvert f+g\rvert^{p-1}$ and use this inequality to find

\displaystyle\begin{aligned}\lvert f+g\rvert^p&=\lvert f+g\rvert\,\lvert f+g\rvert^{p-1}\\&\leq(\lvert f\rvert+\lvert g\rvert)\lvert f+g\rvert^{p-1}\\&=\lvert f\rvert\,\lvert f+g\rvert^{p-1}+\lvert g\rvert\,\lvert f+g\rvert^{p-1}\\&\leq\left(\lvert f\rvert^p+\lvert g\rvert^p\right)^\frac{1}{p}\left(2\lvert f+g\rvert^{q(p-1)}\right)^\frac{1}{q}\\&\leq\left(\lvert f\rvert^p+\lvert g\rvert^p\right)^\frac{1}{p}2^\frac{1}{q}\lvert f+g\rvert^{p-1}\end{aligned}

Dividing out $\lvert f+g\rvert^{p-1}$ and raising both sides to the $p$th power, we conclude that $\lvert f+g\rvert^p\leq 2^\frac{p}{q}\left(\lvert f\rvert^p+\lvert g\rvert^p\right)$. Thus if both $\lvert f\rvert^p$ and $\lvert g\rvert^p$ are integrable, then so is $\lvert f+g\rvert^p$. Thus $f+g$ must be in $L^p$.

Now we calculate

\displaystyle\begin{aligned}\lVert f+g\rVert_p^p&=\int\lvert f+g\rvert^p\,d\mu\\&\leq\int\lvert f\rvert\,\lvert f+g\rvert^{p-1}\,d\mu+\int\lvert g\rvert\,\lvert f+g\rvert^{p-1}\,d\mu\\&\leq\left(\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{p}\left(\int\lvert f+g\rvert^{q(p-1)}\,d\mu\right)^\frac{1}{q}+\left(\int\lvert g\rvert^p\,d\mu\right)^\frac{1}{p}\left(\int\lvert f+g\rvert^{q(p-1)}\,d\mu\right)^\frac{1}{q}\\&\leq\left(\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{p}\left(\left(\int\lvert f+g\rvert^p\,d\mu\right)^\frac{1}{p}\right)^\frac{p}{q}+\left(\int\lvert g\rvert^p\,d\mu\right)^\frac{1}{p}\left(\left(\int\lvert f+g\rvert^p\,d\mu\right)^\frac{1}{p}\right)^\frac{p}{q}\\&=\left(\lVert f\rVert_p+\lVert g\rVert_p\right)\left(\lVert f+g\rVert_p\right)^\frac{p}{q}\end{aligned}

Dividing out by $\left(\lVert f+g\rVert_p\right)^\frac{p}{q}$ we find that

$\displaystyle\lVert f+g\rVert_p=\left(\lVert f+g\rVert_p\right)^\frac{p}{p}=\left(\lVert f+g\rVert_p\right)^{p\left(1-\frac{1}{q}\right)}=\left(\lVert f+g\rVert_p\right)^{p-\frac{p}{q}}\leq\lVert f\rVert_p+\lVert g\rVert_p$

This lets us conclude that $L^2$ is a vector space. But we can also verify the triangle identity now. Indeed, if $f$, $g$, and $h$ are all in $L^p$, then Minkowski’s inequality shows us that

$\displaystyle\rho_p(f,g)=\lVert f-g\rVert_p\leq\lVert f-h\rVert_p+\lVert h-g\rVert_p=\rho_p(f,h)+\rho_p(h,g)$

which is exactly the triangle inequality we want. Thus $\lVert\cdot\rVert_p$ is a norm, and $L^p$ is a normed vector space.

August 27, 2010 Posted by | Analysis, Measure Theory | 2 Comments

Hölder’s Inequality

We’ve seen the space of integrable functions on a measure space $X$, which we called $L^1$ or $L^1(\mu)$. We’ve seen that this gives us a complete normed vector space — a Banach space. This is what we’d like to generalize.

Given a real number $p>1$, we define the space $L^p$ or $L^p(\mu)$ to be the collection of all measurable functions $f$ for which $\lvert f\rvert^p$ is integrable. As in the case of $L^1$, we identify two functions if they’re equal $\mu$-almost everywhere.

It will turn out that these are Banach spaces. We define the $L^p$ norm

$\displaystyle\lVert f\rVert_p=\left(\int\lvert f\rvert^p\,d\mu\right)^{\frac{1}{p}}$

and we write $\rho_p(f,g)=\lVert f-g\rVert_p$ to define a metric. This is clearly non-negative, and we see that $\rho_p(f,g)=0$ if and only if $f=g$ $\mu$-a.e., just as before. It’s also clear that $\lVert cf\rVert_p=\lvert c\rvert\lVert f\rVert_p$. What we need to work to check is the triangle inequality. It’s also not quite so apparent a problem, but we actually don’t know yet that this is a vector space at all! That is, how do we know that $\lvert f-g\rvert^p$ is integrable if $\lvert f\rvert^p$ and $\lvert g\rvert^p$ are?

As a first step in this direction, we prove Hölder’s inequality: if $p$ and $q$ are real numbers greater than $1$ such that $\frac{1}{p}+\frac{1}{q}=1$, and if $f\in L^p$ and $g\in L^q$, then the product $fg\in L^1$ and $\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$. To see this, we will use the function $\phi$ defined for all positive real numbers by

$\displaystyle\phi(t)=\frac{t^p}{p}+\frac{t^{-q}}{q}$

Differentiating, we see that $\phi'(t)=t^{p-1}-t^{-q-1}=t^{-q-1}(t^{p+q}-1)$, so the only (positive) critical point of $\phi$ is $1$. Since the limit as $t$ approaches $0$ and $\infty$ are both positive infinite, $t=1$ must be a local minimum. That is

$\displaystyle\frac{t^p}{p}+\frac{t^{-q}}{q}=\phi(t)\geq\phi(1)=\frac{1}{p}+\frac{1}{q}=1$

For any two real numbers $a$ and $b$, we can consider the value

$\displaystyle t=\frac{a^\frac{1}{q}}{b^\frac{1}{p}}$

and it follows that

\displaystyle\begin{aligned}1&\leq\phi(t)\\&=\frac{t^p}{p}+\frac{t^{-q}}{q}\\&=\frac{\frac{a^\frac{p}{q}}{b}}{p}+\frac{\frac{a^{-1}}{b^{-\frac{q}{p}}}}{q}\\&=\frac{a^\frac{p}{q}}{bp}+\frac{b^\frac{q}{p}}{aq}\\&=\frac{a^{p-1}}{bp}+\frac{b^{q-1}}{aq}\end{aligned}

and thus

$\displaystyle ab\leq\frac{a^p}{p}+\frac{b^q}{q}$

which is clearly also true even if we allow $a$ or $b$ to be zero. This is known as “Young’s inequality”.

Okay, so now we can turn to the theorem itself. If either $\lVert f\rVert_p=0$ or $\lVert g\rVert_q=0$, the inequality clearly holds. Otherwise, we define

$\displaystyle a=\frac{\lvert f\rvert}{\lVert f\rVert_p}\qquad b=\frac{\lvert g\rvert}{\lVert g\rVert_q}$

we can plug these into the above inequality to find

\displaystyle\begin{aligned}\frac{\lvert fg\rvert}{\lVert f\rVert_p\lVert g\rVert_q}&=\frac{\lvert f\rvert}{\lVert f\rVert_p}\frac{\lvert g\rvert}{\lVert g\rVert_q}\\&\leq\frac{\lvert f\rvert^p}{p\lVert f\rVert_p^p}+\frac{\lvert g\rvert^q}{q\lVert g\rVert_q^q}\\&=\frac{1}{p}\frac{\lvert f\rvert^p}{\int\lvert f\rvert^p\,d\mu}+\frac{1}{q}\frac{\lvert g\rvert^q}{\int\lvert g\rvert^q\,d\mu}\end{aligned}

Since the measurability of $f$ and $g$ implies that of $\lvert fg\rvert$, and the right hand side of this inequality is integrable, we conclude that $\lvert fg\rvert$ is integrable. If we integrate, we find

$\displaystyle\frac{\lVert fg\rVert_1}{\lVert f\rVert_p\lVert g\rVert_q}\leq\frac{1}{p}\frac{\int\lvert f\rvert^p\,d\mu}{\int\lvert f\rvert^p\,d\mu}+\frac{1}{q}\frac{\int\lvert g\rvert^q\,d\mu}{\int\lvert g\rvert^q\,d\mu}=\frac{1}{p}+\frac{1}{q}=1$

and Hölder’s inequality follows.

The condition relating $p$ and $q$ is very common in this discussion, so we will say that such a pair of real numbers are “Hölder conjugates” of each other. Given $p$, the Hölder conjugate $q$ is uniquely defined by $q=\frac{p}{p-1}$, which is a strictly decreasing function sending $(1,\infty)$ to itself (with order reversed, of course). The fact that this function has a (unique) fixed point at $2$ will be important. In particular, we will see that this norm is associated with an inner product on $L^2$, and that Hölder’s inequality actually implies the Cauchy-Schwarz inequality!

August 26, 2010 Posted by | Analysis, Measure Theory | 6 Comments

The Measure Algebra of the Unit Interval

Let $Y$ be the unit interval $[0,1]$, let $\mathcal{T}$ be the class of Borel sets on $Y$, and let $\nu$ be Lebesgue measure. If $\{\mathcal{Q}_n\}$ is a sequence of partitions of the maximal element $Y$ of the measure algebra $(\mathcal{T},\nu)$ into intervals, and if the limit of the sequence of norms $\lvert\mathcal{Q}_n\rvert$ is zero, then $\{\mathcal{Q}_n\}$ is dense.

If $\epsilon$ is a positive number, then we can find some $n$ so that $\lvert\mathcal{Q}_n\rvert<\frac{\epsilon}{2}$. If $E$ is a subinterval of $Y$, then let $E_1\in\mathcal{Q}_n$ be the unique interval containing the left endpoint of $E$. If $E_1$ also contains the right endpoint, then we can stop. Otherwise, let $E_2$ be the next interval of $\mathcal{Q}_n$ to the right of $E_1$, and keep going (at most a finite number of steps) until we get to an interval $E_k$ containing the right endpoint of $E$. The union $U$ of all the $\{E_i\}$ can overshoot $E$ by at most $\frac{\epsilon}{2}$ on the left and the same amount on the right, and so $\rho(E,U)<\epsilon$. Thus any interval can be approximated arbitrarily closely by some partition in the sequence. The general result follows, because we can always find a finite collection of intervals whose union is arbitrarily close to any given Borel set.

Now, say that every separable, non-atomic, normalized measure algebra $(\mathcal{S},\mu)$ is isomorphic to the measure ring $(\mathcal{T},\nu)$.

Since the metric space $\mathfrak{S}(\mu)$ has a diameter of $1$ — no two sets can differ by more than $X$, and $\mu(X)=1$ — we can find a dense sequence $\{E_k\}$ in the space. For each $n$ we can consider sets of the form

$\displaystyle\bigcap\limits_{i=1}^n A_i$

where either $A_i=E_i$ or $A_i=X\setminus E_i$. The collection of all such sets for a given $n$ is a partition $\mathcal{P}_n$. It should be clear that the sequence $\{\mathcal{P}_n\}$ is decreasing, and the fact that $\{E_n\}$ is dense implies that the sequence of partitions is also dense. We thus conclude that $\lim\limits_{n\to\infty}\lvert\mathcal{P}_n\rvert=0$.

The first partition $\mathcal{P}_1$ has two elements $E_1$ and $X\setminus E_1$. We can define $t(E_1)=[0,\mu(E_1)]$ and $t(X\setminus E_1)=(\mu(E_1),1]$, so that $\nu(t(E_1))=\mu(E_1)$ and $\nu(t(X\setminus E_1))=\mu(X\setminus E_1)$. This gives us a partition $\mathcal{Q}_1$ of $Y$ that reflects the structure of $\mathcal{P}_1$. Similarly, we can carve up each of these intervals so that the resulting partition $\mathcal{Q}_2$ reflects the structure of $\mathcal{P}_2$. And we can continue, each time subdividing $\mathcal{Q}_n$ into smaller intervals so that the next partition $\mathcal{Q}_{n+1}$ reflects the structure of $\mathcal{P}_{n+1}$. Since this correspondence preserves measures, it follows that $\lim\limits_{n\to\infty}\lvert\mathcal{Q}_n\rvert=0$, and the above result then shows that the sequence $\{\mathcal{Q}_n\}$ is dense.

Now, we can extend $t$ from partition elements occurring in $\{\mathcal{P}_n\}$ to finite unions of such elements by sending such a finite union to the finite union of corresponding elements of $\{\mathcal{Q}_n\}$. This gives an isometry from a dense subset of $\mathfrak{S}(\mu)$ to a dense subset of $\mathfrak{T}(\nu)$. Thus we can uniquely extend it to an isometry $\bar{t}:\mathfrak{S}(\mu)\to\mathfrak{T}(\nu)$. Since $t$ preserves unions and differences, and since these operations are uniformly continuous, it follows that $\bar{t}$ gives us an isomorphism of measure algebras.

August 25, 2010

Partitions in Measure Algebras

Let $(\mathcal{S},\mu)$ be a totally finite measure algebra, and write $X$ for the maximal element. Without loss of generality, we can assume that $\mu$ is normalized so that $\mu(X)=1$.

We define a “partition” $\mathcal{P}$ of an element $E\subseteq\mathcal{S}$ to be a finite set of “disjoint” elements of $\mathcal{S}$ whose “union” is $E$. Remember, of course, that the elements of $\mathcal{S}$ are not (necessarily) sets, so the set language is suggestive, but not necessarily literal. That is, if $\mathcal{P}=\{E_1,\dots,E_k\}$ then $E_i\cap E_j=\emptyset$ and

$\displaystyle E=\bigcup\limits_{i=1}^kE_i$

The “norm” $\lvert\mathcal{P}\rvert$ of a partition $\mathcal{P}$ is the maximum of the numbers $\{\mu(E_i)\}$. If $\mathcal{P}=\{E_1,\dots,E_k\}$ is a partition of $E$ and if $F\subseteq E$ is any element of $\mathcal{S}$ below $E$, then $\mathcal{P}\cap F=\{E_1\cap F,\dots,E_k\cap F\}$ is a partition of $F$.

If $\mathcal{P}_1$ and $\mathcal{P}_2$ are partitions, then we write $\mathcal{P}_1\leq\mathcal{P}_2$ if each element in $\mathcal{P}_1$ is contained in an element of $\mathcal{P}_2$. We say that a sequence of partitions is “decreasing” if $\mathcal{P}_{n+1}\leq\mathcal{P}_n$ for each $n$. A sequence of partitions is “dense” if for every $E\in\mathcal{S}$ and every positive number $\epsilon$ there is some $n$ and an element $E_0\in\mathcal{S}$ so that $\rho(E,E_0)<\epsilon$, and $E_0$ is exactly the union of some elements in $\mathcal{P}_n$. That is, we can use the elements in a fine enough partition in the sequence to approximate any element of $\mathcal{S}$ as closely as we want.

Now, if $(\mathcal{S},\mu)$ is a totally finite, non-atomic measure algebra, and if $\{\mathcal{P}_n\}$ is a dense, decreasing sequence of partitions of $X$, then $\lim\limits_{n\to\infty}\lvert\mathcal{P}_n\rvert=0$. Indeed, the sequence of norms $\{\lvert\mathcal{P}_n\rvert\}$ is monotonic and bounded in the interval $[0,1]$, and so it must have a limit. We will assume that this limit is some positive number $\delta>0$, and find a contradiction.

So if $\mathcal{P}_1=\{E_1,\dots,E_k\}$ then at least one of the $E_i$ must be big enough that $\lvert\mathcal{P}_n\cap E_i\rvert\geq\delta$ for all $n$. Otherwise the sequence of norms would descend below $\delta$ and that couldn’t be the limit. Let $F_1$ be just such an element, and consider the sequence $\{\mathcal{P}\cap F_1\}$ of partitions of $F_1$. The same argument is just as true, and we find another element $F_2\subseteq F_1$ from the partition $\mathcal{P}_2$, and so on.

Now, let $F$ be the intersection of the sequence $\{F_n\}$. By assumption, each of the $F_n$ has $\mu(F_n)\geq\delta$, and so $\mu(F)\geq\delta$ as well. Since $(\mathcal{S},\mu)$ is non-atomic, $F$ can’t be an atom, and so there must be an $F_0\subseteq F$ with $0<\mu(F_0)<\mu(F)$. This element must be either contained in or disjoint from each element of each partition $\mathcal{P}_n$.

We can take $\epsilon$ smaller than either $\mu(F_0)$ or $\mu(F)-\mu(F_0)$. Now no set made up of the union of any elements of any partition $\mathcal{P}_n$ can have a distance less than $\epsilon$ from $F_0$. This shows that the sequence of partitions cannot be dense, which is the contradiction we were looking for. Thus the limit of the sequence of norms is zero.

August 24, 2010 Posted by | Analysis, Measure Theory | 1 Comment

Stone Spaces

The Stone space functor we’ve been working with sends Boolean algebras to topological spaces. Specifically, it sends them to compact Hausdorff spaces. There’s another functor floating around, of course, though it might not be the one you expect.

The clue is in our extended result. Given a topological space $X$ we define $S(X)$ to be the Boolean algebra of all clopen subsets. This functor is contravariant — given a continuous map $f:X\to Y$, we get a homomorphism of Boolean algebras $S(f)$ sending the clopen set $Z\subseteq Y$ to its preimage $f^{-1}(Z)\subseteq X$. It’s straightforward to see that this preimage is clopen. Another surprise is that this is known as the “Stone functor”, not to be confused with the Stone space functor $S(\mathcal{B})$.

So what happens when we put these two functors together? If we start with a Boolean algebra $\mathcal{B}$ and build its Stone space $S(\mathcal{B})$, then the Stone functor applied to this space gives us a Boolean algebra $S(S(\mathcal{B}))$. This is, by construction, isomorphic to $\mathcal{B}$ itself. Thus the category $\mathbf{Bool}$ is contravariantly equivalent to some subcategory $\mathbf{Stone}$ of $\mathbf{CHaus}$. But which compact Hausdorff spaces arise as the Stone spaces of Boolean algebras?

Look at the other composite; starting with a topological space $X$, we find the Boolean algebra $S(X)$ of its clopen subsets, and then the Stone space $S(S(X))$ of this Boolean algebra. We also get a function $X\to S(S(X))$. For each point $x\in X$ we define the Boolean algebra homomorphism $\lambda_x:S(X)\to\mathcal{B}_0$ that sends a clopen set $C\subseteq X$ to $1$ if and only if $x\in C$. We can see that this is a continuous map by checking that the preimage of any basic set is open. Indeed, a basic set of $S(S(X))$ is $s(C)$ for some clopen set $C\subseteq X$. That is, $\{\lambda\in S(S(X))\vert\lambda(C)=1\}$. Which functions of the form $\lambda_x$ are in $s(C)$? Exactly those for which $x\in C$. Since $C$ is clopen, this preimage is open.

Two points $x_1$ and $x_2$ are sent to the same function $\lambda_{x_1}=\lambda_{x_2}$ if and only if every clopen set containing $x_1$ also contains $x_2$, and vice versa. That is, $x_1$ and $x_2$ must be in the same connected component. Indeed, if they were in different connected components, then there would be some clopen containing one but not the other. Conversely, if there is a clopen that contains one but not the other they can’t be in the same connected component. Thus this map $X\to S(S(X))$ collapses all the connected components of $X$ into points of $S(S(X))$.

If this map $X\to S(S(X))$ is a homeomorphism, then no two points of $X$ are in the same connected component. Thus each singleton $\{x\}\subseteq X$ is a connected component, and we call the space “totally disconnected”. Clearly, such a space is in the image of the Stone space functor. On the other hand, if $X=S(\mathcal{B})$, then $S(S(X))=S(S(S(\mathcal{B})))\cong S(\mathcal{B})=X$, and so this is both a necessary and a sufficient condition. Thus the “Stone spaces” form the full subcategory of $\mathbf{CHaus}$, consisting of the totally disconnected compact Hausdorff spaces. Stone’s representation theorem shows us that this category is equivalent to the dual of the category of Boolean algebras.

As a side note: I’d intended to cover the Stone-Čech compactification, but none of the references I have at hand actually cover the details. There’s a certain level below which everyone seems to simply assert certain facts and take them as given, and I can’t seem to reconstruct them myself.

August 23, 2010