# The Unapologetic Mathematician

## Cauchy’s Condition

We defined the real numbers to be a complete uniform space, meaning that limits of sequences are convergent if and only if they are Cauchy. Let’s write these two out in full:

• A sequence $a_n$ is convergent if there is some $L$ so that for every $\epsilon$ there is an $N$ such that $n>N$ implies $|a_n-L|<\epsilon$.
• A sequence $a_n$ is Cauchy if for every $\epsilon$ there is an $N$ such that $m>N$ and $n>N$ implies $|a_m-a_n|<\epsilon$.

See how similar the two definitions are. Convergent means that the points of the sequence are getting closer and closer to some fixed $L$. Cauchy means that the points of the sequence are getting closer to each other.

Now there’s no reason we can’t try the same thing when we’re taking the limit of a function at $\infty$. In fact, the definition of convergence of such a limit is already pretty close to the above definition. How can we translate the Cauchy condition? Simple. We just require that for every $\epsilon>0$ there exist some $R$ so that for any two points $x,y>R$ we have $|f(x)-f(y)|<\epsilon$.

So let’s consider a function $f$ defined in the ray $\left[a,\infty\right)$. If the limit $\lim\limits_{x\rightarrow\infty}f(x)$ exists, with value $L$, then for every $\epsilon>0$ there is an $R$ so that $x>R$ implies $|f(x)-L|<\frac{\epsilon}{2}$. Then taking $y>R$ as well, we see that

$|f(x)-f(y)|=|(f(x)-L)-(f(y)-L)|\leq|f(x)-L|+|f(y)-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

and so the Cauchy condition holds.

Now let’s assume that the Cauchy condition holds. Define the sequence $a_n=f(a+n)$. This is now a Cauchy sequence, and so it converges to a limit $A$, which I assert is also the limit of $f$. Given an $\epsilon>0$, choose an $R$ so that

• $|f(x)-f(y)|<\frac{\epsilon}{2}$ for any two points $x$ and $y$ above $R$
• $|a_n-A|<\frac{\epsilon}{2}$ whenever $a+n\geq B$

Just take a $B$ for each condition, and go with the larger one. In fact, we may as well round $B$ up so that $B=a+N$ for some natural number $N$. Then for any $b>B$ we have

$|f(b)-A|=|(f(b)-f(B))+(f(B)-A)|\leq|f(b)-f(B)|+|a_N-A|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

and so the limit at infinity exists.

In the particular case of an improper integral, we have $I(b)=\int_a^bf\,d\alpha$. Then $I(c)-I(b)=\int_b^cf\,d\alpha$. Our condition then reads:

For every $\epsilon>0$ there is a $B$ so that $c>b>B$ implies $\left|\int_b^cfd\alpha\right|<\epsilon$.

April 23, 2008 - Posted by | Analysis, Calculus