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.

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.