# Uniformly convergent sequence proof.

Prove that if $(f_k)$ is a uniformly convergent sequence of continuous real-valued functions on a compact domain $D\subseteq \mathbb{R}$, then there is some $M\geq 0$ such that $\left|f_k(x)\right|\leq M$ for every $x\in D$ and for every $k\in \mathbb{N}$.

My response: Basically, I am trying to show that uniform convergence on a compact domain implies uniform boundedness. Let $f(x)$ be the limiting function. Then I know that $\lim_{k\to\infty} \sup_{x \in D} | f_k (x) - f(x) | = 0$. Also, I know that $f$ is continuous, therefore it attains an absolute maximum $\in D$. How can I apply these two things to prove it?

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Since the sequence $\|f_k-f\|_\infty$ is convergent, it is bounded : $\|f_k-f\|_\infty\leq A$. On the other side, $D$ is compact and $f$ continuous so the function $f$ is also bounded : $\|f\|_\infty\leq M$. Finally, because $\|.\|_\infty$ is norm or more precisely due to the triangle inequality, it implies that $|\|f_k\|_\infty-\|f\|_\infty|\leq \|f_k-f\|_\infty$. Hence you can bound $\|f_k\|_\infty$ by a constant which does not depend on $k$. – Bebop Dec 9 '12 at 3:51

You are very much on the right track. So I'll try to bump you the rest of the way there.

We want to find a bound on all $f_k$ simultaneously.

It is not too hard to simultaneously bound any finite collection of our functions. This means we can control any finite initial segment of our sequence fairly easily.

As you have noted, $f$ is also continuous and hence has a maximum $M$.

Then by uniform convergence have some $N$ large so that for $n\geq N$ we know that $$\lvert f(x) - f_n(x)\rvert < 1$$ for all $x$.

This gives us a bound on the tail of the sequence, they can be no bigger than $M + 1$.

So now can you combine these controls over the tail and initial segment of our sequence to finish the proof?

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I think this is quite obvious. Noted that $D$ is compact and $f$ is continuous so $f_k(D)$ is also compact for every $k$ and hence it is bounded for every $k\in\mathbb{N}$ and we can just take $M=\sup\bigcup f_k(D)$ where $M \ne +\infty$ as every $f_k(D)$ are bounded.

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The point is exactly about the supremum : $\sup\cup f_k(D)$. It is certainly not clear that this is not $\infty$. Look at my comment or Deven's answer. – Bebop Dec 9 '12 at 11:22
@Bebop But every $f_k(D)$ is bounded, can it still possible equal to $\infty$. If there exist $f_n(p)=\infty\implies f$ is unbounded. – Mathematics Dec 9 '12 at 11:24
@Bebop I am quite confused why i cannot prove in this way, can you tell about the mistake i made in the proof above? – Mathematics Dec 9 '12 at 11:33
The problem is that just because each individual $f_n$ is bounded doesn't mean they are uniformly bounded, this is the reason we are required to say they uniformly converge. For example $f_n = n$ then each $f_n(D)$ is bounded but $\sup \bigcup f_k(D) = \infty$. – Deven Ware Dec 10 '12 at 5:25
We need $D$ to be compact so that the functions will be bounded. For example put $(0,1)$ and $f(x) = \frac{1}{x}$. This screws us up, because now we can just put this at the start of our sequence and not mess up any convergence but the result is now false. – Deven Ware Dec 10 '12 at 6:12