Prove $(C_X,||.||)$ ,where $||.||$ is the maximum norm and X is compact, is complete.

The following proof was given. It is the one I am striving to understand:

Let $(f_n)$ be a Cauchy sequence: $\forall\epsilon>0\exists N\in\mathbb{N}:n,m\geqslant N\implies ||f_m-f_n||<\epsilon$

$\forall t\in X$

$0\leqslant |f_n(t)-f_m(t)|\leqslant \max_{x\in X}|f_n(x)-f_m(x)|\to 0$ as $m,n\to\infty$

$\forall t\in X\:, (f_n(t))_{n\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$.

Then $(f_n)_n\to f$ uniformly then $f$ is continuous.

This is how the proof was handed to me. I think I can fill the gaps but I would need someone to back me on that.

So first the author considers a Cauchy sequence and assumes it converges in $C(X)$ Then it arrives to the following inequality: $0\leqslant |f_n(t)-f_m(t)|\leqslant \max_{x\in X}|f_n(x)-f_m(x)|\to 0$ as $m,n\to\infty$ since the it assumed $\max_{x\in X}|f_n(x)-f_m(x)|\to 0$ then $|f_n(t)-f_m(t)|\to 0$ So the convergence in $C(X)$ verifies that the same Cauchy sequence converges in $\mathbb{R}$ that is by assumption complete with the usual topology. Since $X$ is compact then $f_n$ converges uniformly in $\mathbb{R}$ and it converges to a continuous function. Therefore it converges in $C(X)$ proving the latter is complete.


Is this the reasoning behind the proof?

Thanks in advance!

  • 1
    $\begingroup$ The author doesn't assume that an arbitrary Cauchy sequence converges. The author is trying to prove that it does. $\endgroup$
    – parsiad
    Feb 15 '19 at 21:56
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    $\begingroup$ No, you can't take a Cauchy sequence in the space you want to prove is complete and then assume that it converges! You must prove that it converges. $\endgroup$ Feb 15 '19 at 21:56
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    $\begingroup$ Since $f_n(x)$ is a real number for any $n$, $f_n(x)$ is a sequence of real numbers. Because we have shown that it's a Cauchy sequence and $\mathbb{R}$ is complete, $f_n(x)$ must converge to some real number that we can call it $f(x)$. $\endgroup$ Feb 15 '19 at 21:58
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    $\begingroup$ You showed that for each $t$, $(f_n(t))_n$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete, this sequence converges to some point, call it $f(t)$. $\endgroup$
    – parsiad
    Feb 15 '19 at 21:58
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    $\begingroup$ The author uses $$0\leqslant |f_n(t)-f_m(t)|\leqslant \max_{x\in X}|f_n(x)-f_m(x)|\to 0$$ to show that $f_n(x)$ is a Cauchy sequence in $\mathbb{R}$. $\endgroup$ Feb 15 '19 at 21:59

Let $X$ be a compact space. Let $C_{X}$ be the space of functions mapping from $X$ to $\mathbb{R}$. For any function $f$ in $C_{X}$, define $$ \left\Vert f\right\Vert \equiv\max_{x}\left|f(x)\right|. $$ This is called the maximum norm. We would like to prove the following:

Theorem. $(C_{X},\Vert\cdot\Vert)$ is complete.

Definition. A space $(Y, |\cdot|)$ is complete if any sequence $(y_n)_n$ of points in $Y$ which is Cauchy with respect to the norm $|\cdot|$ converges to some point $y$ in $Y$.

Remark. Don't be confused by the terminology "points" here. Since $C_{X}$ is a space of functions, its points are functions.

Proof. Start with an arbitrary Cauchy sequence $(f_{n})_{n}$ in $C_{X}$. Let $t$ be an arbitrary point in $X$. Note that $$ \left|f_{n}(t)-f_{m}(t)\right|\leq\max_{t}\left|f_{n}(t)-f_{m}(t)\right|=\left\Vert f_{n}-f_{m}\right\Vert .\tag{1} $$ Let $\epsilon$ be a positive constant and pick $N$ such that $\Vert f_{n}-f_{m}\Vert<\epsilon$ for all $n,m\geq N$. We can do this because we assumed the sequence $(f_{n})_{n}$ was Cauchy. By the above inequality, it follows that $|f_{n}(t)-f_{m}(t)|$ is also strictly less than $\epsilon$. Therefore, the sequence $(f_{n}(t))_{n}$ is also Cauchy.

Remark. Note that the sequences $(f_{n})_{n}$ and $(f_{n}(t))_{n}$ are different! One is a sequence of functions and one is a sequence of numbers in $\mathbb{R}$.

Now, since $\mathbb{R}$ is complete, it follows that $(f_{n}(t))_{n}$ converges to some real number. Let's call that real number $f(t)$. Since $t$ was arbitrary, we have essentially defined a new function, $f:X\rightarrow\mathbb{R}$.

Lastly, let's make sure that $f_{n}$ converges to this new function $f$ with respect to the maximum norm. Taking limits with respect to $m$ in the inequality (1), $$ \left|f_{n}(t)-f(t)\right|=\lim_{m}\left|f_{n}(t)-f_{m}(t)\right|\leq\lim_{m}\left\Vert f_{n}-f_{m}\right\Vert =\left\Vert f_{n}-f\right\Vert . $$ In the above, we have used the fact that limits and continuous functions commute and norms are continuous. This implies that $f$ is continuous since if a sequence of continuous functions converge "uniformly" (i.e., with respect to the maximum norm) to some function, that function must be continuous.

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    $\begingroup$ I can't upvote your answer because I have used all my 40 votes for today, but good explanation. As a side note, there is nothing special about $\mathbb{R}$. We can have any Banach space instead of $\mathbb{R}$. Most importantly, the same proof works if we replace $\mathbb{R}$ with $\mathbb{R}^n$ or $\mathbb{C}^n$. $\endgroup$ Feb 15 '19 at 22:24
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    $\begingroup$ @stressedout: +1. You are right. I was trying to make it as accessible as possible. $\endgroup$
    – parsiad
    Feb 15 '19 at 22:28

This is supposed to be a sidenote to parsad's answer. You may not understand it now because some of the terms are new for you, but you can return to it in future.

I want to add that there is nothing special about $\mathbb{R}$. As long as $Y$ is a complete vector space, like $\mathbb{R}^n$, $\mathbb{C}^n$ or even more complicated spaces like $\mathcal{L}^p(X)$, the space $C(X,Y)= \{f \mid f:X\xrightarrow{continuous} Y \}$ with the following norm is complete:

$$\|f\|=\max_{x \in X}\|f(x)\|_Y$$

where $\|\cdot\|_Y$ denotes the norm of $Y$. In our special case, $Y=\mathbb{R}$ and $\|\cdot\|_Y=| \cdot |$ (the absolute value).

The proof is exactly the same. Let $(f_n)_{n=1}^{\infty}$ be a Cauchy sequence in $C(X,Y)$, then $(f_n(x))_{n=1}^{\infty}$ is Cauchy in $Y$ because

$$0 \leq \|f_n(x) - f_m(x)\|_Y \leq \max_{x\in X}\|f_n(x)-f_m(x)\|_Y=\|f_n-f_m\|\to0$$

Since $Y$ is complete, for each $x\in X$ we can define $\lim_{n\to\infty}f_n(x):=f(x)$. Now to show that $f(x)$ is continuous, consider the following inequality:

$$\|f(x)-f(y)\|_Y \leq \|f(x)-f_N(x)\|_Y+\|f_N(x)-f_N(y)\|_Y+\|f_N(y)-f(y)\|_Y \hspace{10px} (\star)$$

Given $\epsilon > 0$, there exists $N_1$ and $N_2$ such that

$$n \geq N_1 \implies \|f(x)-f_n(x)\|_Y < \epsilon/3$$ and $$n \geq N_2 \implies \|f(y)-f_n(y)\|_Y<\epsilon/3$$

Take $N=\max(N_1,N_2)$. Then since $X$ is compact, $f_N$ is uniformly continuous, and we can find a neighborhood such that $\forall x,y: \|f_N(x)-f_N(y)\|_Y < \epsilon/3$.

Combining our inequalities for $N=\max(N_1,N_2)$ in $(\star)$, we have that $\|f(x)-f(y)\|_Y < \epsilon$ which proves that $f$ is continuous and hence, $f \in C(X,Y)$. Q.E.D.

  • $\begingroup$ as Alex East point out, the compactness is for the finite property in definition of norm, and then no need using it in proof. $\endgroup$
    – ydhhat
    Oct 31 '20 at 12:08

This is a supplement to the answer of @stressedout and @parsiad. Although both proofs are sound they leave out one key point, the justification of compactness in the assumptions. The compactness of the space is not needed for the proof of continuity of the limit function as we only need to prove: $\forall x : \forall \varepsilon \in K : \exists U$ open such that $\left \| f(x) - f(y) \right \| < \varepsilon, \forall y \in U$. So the $x$ in this context is fixed and "normal" continuity (which we have from the get go), will suffice. You can see this done for instance in the proof of the uniform limit theorem: https://en.wikipedia.org/wiki/Uniform_limit_theorem

It's more basic than that. If you take $C(X)$ where $X$ is not compact with the supremum norm you will certainly not (in general) have finite norm values on all your functions in $C(X)$. Therefore we consider a compact space X since the extreme value theorem ensures that the norms are finite: http://mathonline.wikidot.com/the-extreme-value-theorem-for-continuous-functions-on-compac


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