Let $X=C(I)$ with the sup norm $\|\cdot\|_\infty$, where $I=[0,1]$. Let $K$ be a Volterra integral operator:
$$Kx(t)=\int_0^tK(t,\tau)x(\tau)d\tau, x\in X$$
Show that K: $X\rightarrow X$, is compact
I am having trouble with the $t$ variable present both in $K(t,\cdot)$ and in the integral integration limit. Most examples I've found are easier because they have $K(t,\tau)=1$ or the integral is from 0 to 1.
I have already shown that K is linear,
1)K is bounded:
$|Kf(t)|=|\int_0^tK(t,\tau)f(\tau)d\tau|\le \int_0^t|K(t,\tau)f(\tau)|d\tau \le \int_0^1|K(t,\tau)f(\tau)|d\tau \le\| f\|_{\infty}\int_0^1|K(t,\tau)|d\tau $
Since $K$ is continuous over a compact set, by Weierstrass theorem $\|K(\cdot, \cdot)\|\le M$, with $M$ some real number so $|Kf(t)|=\| f(\tau)\|_{\infty}M $
Then $\|Kf\|_\infty=\sup_t|Kf(t)|\le\| f\|_{\infty}M $ so K is bounded and $\|K\|\le M $
2)$K$ is compact
To show T is compact I have to use Ascoli-Arzelà theorem to the image of the unit ball with rispect to the sup norm: $K(B(0;1))$, so I need to verify the hypotheses of the theorem
$B=B(0;1)=\{f \in C[0,1], \|f\|_\infty <1\}$
i) K(B) is bounded because $\|Kf\|_\infty=\sup_t|Kf(t)|\le\| f\|_{\infty}M \le M \forall f \in B$
ii) K(B) is equicontinuous:
From the uniform continuity of $K$ with respect to the first variable, I have that $\forall \varepsilon >0, \exists \delta > 0$ such that $\forall t_1, t_2 \in [0,1]$, $|K(t_1,\cdot)-K(t_2, \cdot)|\le \varepsilon$...(*)
$|\int_0^tK(t,\tau)f(\tau)d\tau -\int_0^{t_0}K(t_0,\tau)f(\tau)d\tau|$
I don't know how to bound this, if I use the triangle inequality I'd lose the minus sign and then I can't use (*). Can someone shed some light? Is the rest of the proof ok?
hint: The book says that there is a discontinuity line at $\tau = t$ and that in the analogous easier proof with upper integration limit =1 that is given in the book, the continuity of K on the points $(t,\tau)$ for $t<\tau$ was not used. I don't know what they mean or how to use this hint
Edit This is the proof they give in the book for the easier case. Somehow the hint is suggesting that either the proof is the same or there is a slight modification, but I can't figure it out
