# The Unapologetic Mathematician

## 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 -