Show that $f_n\to f$ uniformly

Question

Let $(f_n)$ be a sequence of continuous functions on a compact set $K$ with pointwise and continuous limit $f$. Show that $(f_n)$ converges uniformly to $f$.

My professor gave me a proof, but to be honest, I cannot understand this proof! I cite it therefore:

Quote:

Because of continuity on $K$ it follows that $s_n:=\sup\lvert f_n(x)-f(x)\rvert$ is taken at a $x=x_N$ in $K$ and because of the compactness of $K$ there is a convergent subsequence $y_k=x_{n_k}\to y_*$.

It is true that $$ f_n(y_n)\to f(y_*) $$ because of the continuity of $f_n$ and $f$.

So $$ s_n\to\sup\lvert f(y_*)-f(y_*)\rvert =0. $$ because of the choice of $s_n$ this shows the uniformly convergence.

Do you understand this? I do not.

Answer 1

As others have commented, the statement is false. A very simple counter-example is

$$f_n(x) = \begin{cases} n x & 0\leq x<\frac{1}{n} \\ 2-n x & \frac{1}{n}\leq x<\frac{2}{n} \\ 0 & \frac{2}{n} \leq x \leq 1 \end{cases}.$$

I personally visualize the picture first.

Answer 2

This is false! The mistake in the proof is the claim that $f_n(y_n)\to f(y^*)$. That is true only if you know uniform convergence.

You should have among your bag of examples functions $f_n$ on $[0,1]$ converging non-uniformly to $f=0$.

Answer 3

The proof is wrong as is the statement. You can let $K = [0,1]$ and then let $f_n(x)$ be zero except on $(1/n, 2/n)$ such that $f_n(x)$ has a "hump" of height $1$ between $1/n$ and $2/n$. The pointwise limit is zero but the convergence is not uniform.

The problem with the proof is the statement $f_n(y_n)\to f(y_*)$, which isn't true here. You can see in this example that $f_n(y_n) = 1$ but $f(y_*) = 0$.