C[a,b] with Sup-Norm is complete

Question

I came across many pages claiming to prove that $C[a,b]$ with the Supremum norm: $||f(t)|| = sup_{t\in[a,b]}|f(t)|$ is complete, i.e. that every Cauchy-sequence with elements in $C[a,b]$ converges to a point in $C[a,b]$. Few examples:

• Is the space $C[0,1]$ complete?
• https://mast.queensu.ca/~speicher/Section3.pdf
• https://www.youtube.com/watch?v=g5UjOR1w5NE

All of the examples mentioned above start with showing point-wise convergence to a sequence $(f_n)$ However, none of these examples (as far as I can tell) properly address the issue of uniform convergence of the Cauchy sequence $(f_n)$.

Since $(f_n)$ is Cauchy, $\forall\epsilon>0$ $\;\exists N_1\in\mathbb{N}$ s.t. $||f_n(t)-f_m(t)||<{\frac\epsilon2}$ $\;\forall n,m>N_1$ $\;\forall t\in[a,b]$

$\Rightarrow sup_{t\in[a,b]}|f_n(t)-f_m(t)|<{\frac\epsilon2}$ $\Rightarrow|f_n(t)-f_m(t)|<{\frac\epsilon2}$ $\;\forall t\in[a,b]$.

In particular for each fixed $t_0\in[a,b]$ $|f_n(t_0)-f_m(t_0)|<\epsilon$ $\;\forall n,m>N_1$ Thus, the sequence $(f_n(t_0))$, which is a sequence of real numbers (or complex numbers, that does not matter for the sake of this proof), is also Cauchy.

Since $\mathbb{R}$ is complete, the sequence $(f_n(t_0))$ converges to a point $f(t_0)$, i.e. $\forall\epsilon>0$ $\;\exists N_{t_0}$ s.t. $\forall n>N_{t_0}$ $|f_n(t_0)-f(t_0)|<\epsilon$.

We can construct $f:[a,b]\rightarrow \mathbb{R}$ to obtain point-wise convergence. However, for every $x\in[a,b]$ there is a unique $N_x$, which is a function of $x$, to obtain convergence of the sequence $(f_n(x))$ to $f(x)$.

In order to obtain uniform convergence, one must show that $\forall \epsilon>0 \exists M\in\mathbb{N}$ s.t. $|f_n(t)-f(t)|<\epsilon$ $\forall n>M$ $\forall t\in[a,b]$. Using the triangle inequality:

$|f_n(t)-f(t)|\le |f_n(t)-f_m(t)|+|f_m(t)-f(t)|$

We need to find an $M$ such that the first expression above is less than $\epsilon$ for all $t\in[a,b]$. Choosing $M>N_1$ is obvious, therefore:

$|f_n(t)-f(t)|\le {\frac\epsilon2}+|f_m(t)-f(t)|$

However, what is the selection of $M$ that would make the third expression ($|f_m(t)-f(t)|$) less than ${\frac\epsilon2}$?

Since we proved point-wise convergence, we can find $N_x$ locally, but there is no guarantee that there is a number $N$ such that the inequality is true for all $t\in[a,b]$. Furthermore, proving that the third expression ($|f_m(t)-f(t)|$) is less than ${\frac\epsilon2}$ is the same as proving uniform convergence.

Your help would be very appreciated.

Answer 1

You are wrong if you claim that uniform convergence is not addressed here. The author of that answer wrote “Next you go on to showing that $f_n$ also converges to $f$ in norm”, which, since the norm here is the $\sup$ norm, is exactly the same thing as asserting that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$.

Answer 2

You do not need to select some $M$, the number $N_1$ is completely sufficient. Then you get, with $n,m>N_1$ $$ |f_n(t)-f(t)|\le \fracϵ2+ |f_m(t)-f(t)| $$ as you found. Now the $n$ on the left side is (almost) completely independent of the $m$ on the right side, this inequality is valid for all $m>N_1$, thus also for some $m$ where from the pointwise convergence $$ |f_m(t)-f(t)|<\fracϵ2. $$ Thus the $|f_n(t)-f(t)|<ϵ$, indeed one could sharpen this to $$ |f_n(t)-f(t)|\le \fracϵ2+ \inf_{m>N_1}|f_m(t)-f(t)|=\fracϵ2 $$ as that infimum is obviously zero.