Justify that T is compact I have troubles solving the following problem:
Consider the Hilbert space $H=L_2([0,1],m)$ where m is the Lebesgue measure. Define $K:[0,1]\times [0,1]\to \mathbb{R}$ by:
$K(s,t)=\begin{cases}(1-s)t\ \text{if}\ 0\leq t\leq s\leq 1 \\ (1-t)s\ \text{if}\ 0\leq s<t\leq 1\end{cases}$.
And consider $T\in \mathcal{L}(H,H)$ defined by
$(Tf)(s)=\int_{[0,1]}K(s,t)f(t) dm(t),\ \ \ \ \ s\in[0,1],\ f\in H$
Justify that T is compact.
I am thinking that if I can show that T is of finite rank, then I'd have that T is compact, but I am not sure how to show that T is of finite rank.
 A: Hints: First write down $Tf(s)$ and check that it is a continuous function. Next observe that $K$ is uniformly continuous on $[0,1]\times [0,1]$. By Holder's inequality we have $|Tf(s)-Tf(s')| \le \|f\|_2 (\int_0^{1} |K(s,t)-K(s',t)|^{2}dt)^{1/2}$. Conclude that if $(f_n)$ is a bounded sequence in $L^{2}$ then $(Tf_n)$ is an equi-continuous and uniformly bounded sequence of continuous functions on $[0,1]$. By Arzela -Ascoli Theorem it follows that $(Tf_n)$ has  subsequence which converges uniformly. But then the subsequence also converges in $L^{2}$.
Additional information: It is known that if $K$ is replaced by any square integrable function on the unit square then $T$ is compact.
A: In fact,
\begin{align*}
(Tf)(s)&=\int_0^s(1-s)tf(t)\mathrm{d}t+\int_s^1(1-t)sf(t)\mathrm{d}t
\end{align*}
so
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}s}(Tf)(s)=(1-s)sf(s)+(1-s)sf(s)-\int_0^stf(t)\mathrm{d}t-\int_s^1tf(t)\mathrm{d}t=\int_0^1tf(t)\mathrm{d}t.
\end{align*}
The above computation first works for smooth test functions then extends to $L^2([0,1],\mathrm{d}m)$ and yields (using Cauchy-Schwarz inequality):
\begin{align*}
\left|\frac{\mathrm{d}}{\mathrm{d}s}(Tf)(s)\right|&\leq\int_0^1t|f(t)|\mathrm{d}t\leq\|t\|_{L^2([0,1],\mathrm{d}m)}\|f\|_{L^2([0,1],\mathrm{d}m)}=\frac{1}{3}\|f\|_{L^2([0,1],\mathrm{d}m)}.
\end{align*}
So
\begin{align*}
\left\|\frac{\mathrm{d}}{\mathrm{d}s}(Tf)\right\|_{L^2([0,1],\mathrm{d}m)}&\leq\frac{1}{3}\|f\|_{L^2([0,1],\mathrm{d}m)}.
\end{align*}
This show that $Tf\in H^1([0,1],\mathrm{d}m)$. As the embedding $H^1([0,1],\mathrm{d}m)\subset H^{1-\epsilon}([0,1],\mathrm{d}m)$ is compact for all $\epsilon>0$, we conclude that $T:L^2([0,1],\mathrm{d}m)\to H^{1-\epsilon}([0,1],\mathrm{d}m)$ is compact. In particular, $T:L^2([0,1],\mathrm{d}m)\to L^{2}([0,1],\mathrm{d}m)$ is compact. Actually, a boot-strap argument shows that $T$ is regularizing.
