# Ascoli theorem's proof issues

For future reference; this is a portion of the proof of Ascoli's theorem in Anthony Knapps, Basic Real Analysis.

STATEMENT: Let $(S,d)$ be a compact metric space. If $\left\{f_n\right\}$ is a sequence of scalar-valued functions on $S$ that is equicontinuous at each point of $S$ and point wise bounded on $S$, then

a)$\left\{f_n\right\}$ is a uniformljy equicontinuous and uniformly bounded on $S$.

b)$\left\{f_n\right\}$ has a uniformly convergent subsequence.

Proof:First consider the question of uniform boundedness. Choose, by Corollary 2.39, some $x_n$ in $S$ with $|f(x_n)|=K_n=\sup_{x\in S}|f_n(x)|$ Then choose a subsequence on which the numbers $K_n$ tend to $\sup_nK_n$ in $\mathbb{R}^*$. There will be no loss of generality in assuming that this subsequence is our whole sequence. By compactness of $S$, apply the Bolzano-Weirstrass property given in Theorem 2.36 to find a convergent subsequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$, and let $x_0$ be the limit of this subsequence. By point wise boundedness, find $M_{x_0}$ with $|f_n(x_0)|\leq M_{x_0}$for all $n$. Then choose some $\delta$ of equicontinuity at $x_0$ for $\epsilon=1.$ As soon as $k$ is large enough so that $d(x_{n_k},x_n)<\delta$, we have

$$K_{n_k}=|f_{n_k}(x_{n_k})|\leq |f_{n_k}(x_{n_k})-f_{n_k}(x_0)|+|f_{n_k}(x_0)|<1+M_{x_0}$$

Thus $1+M_{x_0}$ is a uniform bound for the functions $f_n$.

QUESTION: Why can we choose a subsequence that converges to $\sup_n K_n$. It would make sense to find a subsequence that converges to $\limsup_nK_n$. But there doesn't need to have a subsequence that converges to $\sup_nK_n$. I wanted to see if my interpretation is in fact correct of the author's notation in the proof.

• Looks like a minor error, yes, since we can always prepend a huge function before any sequence without changing limit properties. – Ian Dec 9 '14 at 3:29
• Alright thanks. Then that means the rest of the proof is wrong, right, since the $\sup_n K_n$ doesn't need to be bounded by $1+M_{x_0}$. – Enigma Dec 9 '14 at 4:05
• Actually, you can assume that there's a sequence that converges to $\sup_n K_n$. This is not always the case (for a counterexample, take a sequence with this property where $M \equiv \sup_n K_n < \infty$ and then prepend the constant function $2M$ to it). But when it isn't the case, $\sup_n K_n = K_m$ for some $m$, in which case uniform boundedness is given for free. A different way of writing the proof would to be to assume without loss of generality that $\sup_n K_n = \infty$ and then derive a contradiction using the method shown in your proof. – Ian Dec 9 '14 at 4:16
• Well if there is a sequence converging to $M=sup_n K_n$, can't we just directly conclude that $M$ is a uniform bound? – Enigma Dec 9 '14 at 4:19
• Not if $M=\infty$; note that they say that the convergence is in $\mathbb{R}^*$. As I said, a more straightforward method of proof would be "define $M=\sup_{n \in \mathbb{N}} \sup_{x \in S} |f_n(x)|$. If $M<\infty$ then we have uniform boundedness; assume the contrary, then ... contradiction." – Ian Dec 9 '14 at 4:20

## 1 Answer

Technically there may not be a sequence converging to $\sup_n K_n$. This is because $\sup_n K_n$ could be attained at some $K_m$ and then all $K_n$ with $n>m$ could be less than $K_m$. But in this case the uniform bound is just given by $K_m$, so we're done. Hence we can assume that there is a sequence which converges to the supremum (since either the supremum is attained at some $m$ or there is such a sequence).

A more straightforward way to write this proof would be the following.

Define $M=\sup_{n \in \mathbb{N}} \sup_{x \in S} |f_n(x)|$. We seek to show that $M<\infty$. Assume $M=\infty$. Then ..., which is a contradiction. We conclude that $M<\infty$.

Here "..." is the method shown in your post. This approach avoids the technicality in the previous paragraph, because if the supremum is $\infty$ then there must be a sequence which converges, since it cannot be attained at any $m$.