Let $E\subseteq \Bbb R^n$ be measurable and with $|E|>0$ (positive measure), we know $L^2(E)$ is a Hilbert space; so, given $K\in L^2(E\times E)$, we can define the Hilbert-Schmidt operator
$$T_K:L^2(E) \to L^2(E)$$
such that, $\forall f\in L^2(E)$,
$$(T_Kf)(x) := \int_E K(x,y) f(y) dy.$$
Since $L^2(E)$ is a Hilbert space, you can use Schwarts' inequality to determine that $\Vert T_k \Vert_{\mathcal{L}(L^2(E))} \le \Vert K\Vert_{L^2(E\times E)}$, in fact
$$\Vert T_Kf \Vert_{L^2(E)}^2 =\int_E \Bigg( \int_E k(x,y) f(y) dy \Bigg)^2 dx \le \int_E \Bigg( \int_E \big(K(x,y)\big)^2dy \cdot \int_E \big(f(y)\big)^2 dy \Bigg) dx =$$
$$= \int_{E\times E} \big(K(x,y)\big)^2 dxdy \cdot \Vert f \Vert_2^2 = \Vert K \Vert_{L^2(E\times E)}^2 \cdot \Vert f \Vert_{L^2(E)}^2,$$
so that, by definition of norm of an operator, we get that inequality.
My request: does somebody know how to prove that actually equality holds; in other words that, with the generality of the Hilbert-Schmidt operators, we always have
$$\Vert T_k \Vert_{\mathcal{L}(L^2(E))} = \Vert K\Vert_{L^2(E\times E)} .$$
I am not sure that this is true, but in particular cases I seem to be able to show that equality holds and, if not, I can never show that it doesn't.
Best Answer
Conceptually, the $L^2$-norm of $K$ is the Hilbert-Schmidt norm of $T_K$. You are looking for the operator norm of $T_K$, which is usually smaller. In matrices, if you have $$ K=\begin{bmatrix} 1&0\\0&2\end{bmatrix}, $$ then $\|K\|_2=\sqrt5$, while $\|K\|=2$. Below is an adaptation of this idea to your context.
Write $E=E_1\cup E_2$ with $E_1\cap E_2=\emptyset$ and $|E_1|=|E_2|=|E|/2$. Let $$ K(x,y)=1^\vphantom{E}_{E_1\times E_1}+2\,1^\vphantom{E}_{E_2\times E_2}. $$ So $$ T_Kf(x)=\Big(\int_{E_1}f\Big)\,1^\vphantom{E}_{E_1}+2\,\Big(\int_{E_2}f\Big)\,1^\vphantom E_{E_2}. $$ We have \begin{align} \|T_kf\|_2^2&=\Big|\int_{E_1}f\Big|^2\,|E_1|+4\,\Big|\int_{E_2}f\Big|^2\,|E_2| \leq |E_1|^2\,\int_{E_1}|f|^2+4|E_2|^2\,\int_{E_2}|f|^2\\[0.3cm] &=\frac{|E|^2}4\,\bigg[\int_{E_1}|f|^2+4\int_{E_2}|f|^2\bigg] =\frac{|E|^2}4\,\bigg[4\|f\|^2_2-3\int_{E_1}|f|^2\bigg]\\[0.3cm] &\leq |E|^2\,\|f\|_2^2. \end{align} The equality is achieved when $f=\frac{\sqrt2}{|E|^{1/2}}\,\,1_{E_2}$. Thus $\|T_K\|=|E|$.
Meanwhile, $$ \|K\|_2^2=\int_E\int_E |K(x,y)|^2\,dy\,dx =\int_{E\times E}1_{E_1\times E_1}+4\int_{E\times E}1_{E_2\times E_2} =|E_1|^2+4|E_2|^2=\frac{5|E|^2}4. $$ That is, $$ \|K\|_2^2=\frac{\sqrt5\,|E|}2>|E|=\|T_K\|. $$