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Let $V=\mathcal{C}([0,1])$ be the space of all complex valued continuous functions with the norm $||f||_{\infty}=\sup_{x\in[0,1]}|f(x)|$. Then, $V$ with the norm $||\cdot||_{\infty}$ is a Banach space.

I'm having problems to understand some part of the proof. I will proceed proving the previous statement until the question comes.

Given a Cauchy sequence $(f_n)_{n\in \mathbb{N}}\in V$, that's to say, for all $\varepsilon >0$ there exists $N\in \mathbb{N}$ such that for all $n,m \geq N$ we have that $||f_n-f_m|||_\infty \leq \varepsilon$; we need to find a candidate $f$ for the limit of $(f_n)_{n\in \mathbb{N}}$. Fixing $x\in[0,1]$ and noting that for all $n,m\geq N$ we have that $$|f_n(x)-f_m(x)|\leq \sup_{z\in[0,1]}|f_n(z)-f_m(z)|=||f_n-f_m|||_\infty \leq \varepsilon.$$ Thus, $(f_n(x))_{n\in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{C}$ and since $\mathbb{C}$ is complete we know that $\lim_{n\to \infty}f_n(x)$ exists. Then, our candidate will be $f(x):=\lim_{n\to \infty}f_n(x)$.

Now, we want to prove that $f$ is continuous. Fixing $x\in[0,1]$ and letting $\varepsilon >0$ we can choose $N\in \mathbb{N}$ such that for all $n,m\geq N$ we have that $||f_n-f_m|||_\infty \leq \varepsilon/3$. Also, as $f_N\in (f_n)_{n\in \mathbb{N}}\subset V$, for this $\varepsilon >0$ there exists $\delta >0$ such that if $|x-y|\leq \delta$ then $|f_N(x)-f_N(y)|\leq \varepsilon/3$. Furthermore, for any $n,m\geq N$ and $y\in \mathbb{C}$ such that $|x-y|\leq \delta$ we have that \begin{align*} |f(x)-f(y)| =&|f(x)-f_n(x)+f_n(x)-f_N(x)+f_N(x)\\ & -f_N(y)+f_N(y)-f_m(y)+f_m(y)-f(y)| \\ \leq&|f(x)-f_n(x)|+|f_n(x)-f_N(x)|+|f_N(x)-f_N(y)|\\ & +|f_N(y)-f_m(y)|+|f_m(y)-f(y)| \end{align*} I think it's important to do it that way since then $\delta$ could depend on $N$ but not on $n$. The second and the third terms are less or equal that $\varepsilon/3$. However, I don't think I can use the same argument with the fourth, since $y$ may not be in $[0,1]$. The only thing I know is that $|x-y|\leq \delta$. If I prove that the fourth them is less or equal than $\varepsilon/3$, then I know how to finish the prove.

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  • $\begingroup$ How come $y$ may not be in $[0,1]$? $\endgroup$ – Berci Feb 12 at 19:56
  • $\begingroup$ If we can first show $f_n$ uniformly converges to $f$, then the continuity of $f$ follows. $\endgroup$ – Song Feb 12 at 20:07
  • $\begingroup$ @Song this is exactly what is to shown here... so you cannot "prove" a theorem using it. $\endgroup$ – Gono Feb 13 at 6:25
  • $\begingroup$ I believe we can assume $y \in [0,1]$, because we only want to check continuity in the interval $[0,1]$. Also note that $f_n(y)$ is only defined for $y \in [0,1]$! I think this is what @Berci hinted at? $\endgroup$ – red_trumpet Feb 13 at 22:16
  • $\begingroup$ @red_trumpet Yes, exactly. $\endgroup$ – Berci Feb 13 at 22:38
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We'll prove that the $f_n$ converges uniformly to $f$ on $[0,1]$, because this in general a useful result, that uniform convergence of a sequence is in fact equivalent to the sequence being Cauchy in the sup norm. Your desired result will follow because we know that the uniform limit of continuous function is continuous.

Fix $\varepsilon>0$. As you have already shown we know that there exists a $N\in \mathbb N$, such that for all $n,m>N$ and $x\in[0,1]$ we have $$|f_n(x)-f_m(x)|<\varepsilon/2.$$ Now for any $m>N$ and $x\in[0,1]$ we have, that for any $n\in \mathbb N$, $$|f(x)-f_m(x)|\leq|f_m(x)-f_n(x)|+|f(x)-f_n(x)|<\varepsilon/2+|f(x)-f_n(x)|.$$ What is crucial here, is that we can pick any $n$ we like and this inequality holds, which implies that, because of pointwise convergence of $f_n$ to $f$, $$\lim_{n\to\infty}|f(x)-f_m(x)|\leq\lim_{n\to\infty}\varepsilon/2+|f(x)-f_n(x)|=\varepsilon/2<\varepsilon.$$ The LHS is independent of $n$ though, so the limit has no affect. We have thus shown for any $\varepsilon>0$ there exists a $N\in \mathbb N$ such that for all $m>N$ and $x\in[0,1]$ we have $$|f(x)-f_m(x)|<\varepsilon.$$ This is the definition of uniform convergence, which we wanted to show.

Note that you could have directly applied this method to finish your direct proof of continuity. Just take the limit as $n$ and $m$ tend to infinity on both sides of your final inequality.

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  • $\begingroup$ I might be getting something wrong, but shouldn't you use the fact that $[0,1]$ is compact at some point in your proof? OP only showed that for fixed $x \in [0,1]$ there exists $N \in \mathbb{N}$ such that for all $n,m > N$ your inequality holds. $\endgroup$ – Nicolas Feb 13 at 23:07
  • $\begingroup$ @Nicolas You only need compactness to guarantee that the sup norm is actually a norm on the space. The OP's wording is a bit imprecise. He should only have fixed his $x$ after his claim about the existence of $N$, but if you analyse his argument you'll note that his $N$ does not depend on $x$ at all, same as mine doesn't. $\endgroup$ – K.Power Feb 13 at 23:11

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