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Let $K: [a,b] \times [a,b] \to \mathbb{R}$ be a continuous function. The Volterra operator is defined $\mathcal{V} : C ([a,b], \mathbb{R}) \to (C[a,b], \mathbb{R})$

$$ (\mathcal{V}f)(x) := \int_a^b K(x,y)f(y)dy $$

I need to prove that for every bounded sequence of functions $(f_n)_{n \in \mathbb{N}}$ the sequence $(\mathcal{V}f_n)_{n \in \mathbb{N}}$ has a convergent subsequence.

For what I read, what I want is given by Arzela-Ascoli theorem, then I need to verify that $(\mathcal{V}f_n)_{n \in \mathbb{N}}$ is uniformly bounded and equicontinuous.

The set $[a,b] \times [a,b]$ is compact then $\vert K \vert$ have a maximun value $M$ such that $\vert K(x,y) \vert \leq M$ for every $(x,y) \in [a,b] \times [a,b]$, and $(f_n)_{n \in \mathbb{N}}$ is bounded then $\vert f_n(y) \vert \leq c$ for every $n \in \mathbb{N}$ and every $y \in [a,b]$. Then:

$$\vert \mathcal{V}f_n(x) \vert=\lvert \int_a^b K(x,y)f_n(y)dy \rvert \leq \int_a^b \lvert K(x,y) \vert \vert f_n(y) \vert dy \leq Mc(b-a) $$

for every $x \in [a,b]$, then $(\mathcal{V}f_n)_{n \in \mathbb{N}}$ is uniformly bounded. Is this correct?

To prove equicontinuity I need help, please.

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    $\begingroup$ See my point about whether the sequence itself is bounded, or if it is a sequence of bounded functions—the two are not the same. $\endgroup$ – Alex Ortiz Dec 5 '17 at 4:39
  • $\begingroup$ yes, $(f_n)_{n \in \mathbb{N}}$ is a bounded sequence of functions. $\endgroup$ – Mike A. Dec 5 '17 at 6:26
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I have a minor nitpick with the way your question is stated. A sequence of bounded functions is not the same as a bounded sequence of functions. The former (a sequence of bounded functions) is a sequence $\{f_n\}$ such that $|f_n|\le c_n$ for every $n$, but $c_n$ may well depend on $n$. The latter (a bounded sequence of functions) is a sequence $\{g_n\}$ that satisfies $|g_n|\le c$ for every $n$. I assume you mean the latter.

What you have for showing that $\{\mathcal Vf_n\}$ is uniformly bounded looks good. For equicontinuity, write \begin{align*} |\mathcal Vf_n(x+h)-\mathcal Vf_n(x)| &\le \int_a^b |K(x+h,y)-K(x,y)||f_n(y)|\,dy \\ &\le c\int_a^b|K(x+h,y)-K(x,y)|\,dy. \end{align*} Since $[a,b]\times[a,b]$ is compact, $K$ is uniformly continuous, so if $h$ is sufficiently small, $|K(x+h,y)-K(x,y)|<\epsilon/c(b-a)$ for all $y\in[a,b]$. Thus, for all $h$ sufficiently small, $$ c\int_a^b|K(x+h,y)-K(x,y)|\,dy \le c\int_a^b\frac{\epsilon}{c(b-a)}\,y = c\frac{\epsilon}{c(b-a)}(b-a) = \epsilon, $$ which is independent of $n$ and $x$, hence the sequence $\{\mathcal Vf_n\}$ is equicontinuous.

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