Let $k(t, s) = \mathrm e^{t + s^2}$. Show that the integral operator $T : L^2(0,1) \to L^2(0,1)$,
which is defined as
\begin{equation*}
Tf = \int_0^1 k(\cdot, s) f(s) \, \mathrm d s,
\end{equation*}
is bounded.
A Partial answer
We need to show that for all $x \in L^2(0,1)$, $\Vert Tx \Vert_2 \leq M \Vert x \Vert_2$ for some finite $M \geq 0$. Let $f \in L^2(0,1)$.
Then by the ''triangle inequality'' of integration
\begin{align}
\left\Vert Tf(t) \right\Vert_2
%
&= \left\Vert \int_0^1 \mathrm e^{t + s^2} f(s) \, \mathrm d s \right\Vert_2 \\
%
&= \left(\int_0^1 \left|\int_0^1 \mathrm e^{t + s^2} f(s) \, \mathrm d s \right|^2 \,\mathrm dt \right)^{1/2} \\
%
&\leq\left(\int_0^1 \left(\int_0^1 \left|\mathrm e^{t + s^2} f(s)\right| \, \mathrm d s \right)^2 \,\mathrm dt \right)^{1/2} \\
%
&=\left(\int_0^1 \left( \int_0^1 \Big|\mathrm e^{t + s^2}\Big|\Big| f(s)\Big| \, \mathrm d s \right)^2 \,\mathrm dt \right)^{1/2} \,.
\end{align}
Letting $M_1 = \big|\mathrm e^{t + 1}\big| = \mathrm e^{t + 1}$ and $M_2 = \mathrm e^{2}$ (as we are operating on $I = (0,1)$), we can conclude that
\begin{align}
\left\Vert Tf(t) \right\Vert_2
&\leq \left(\int_0^1 \left( \int_0^1 M_1\Big| f(s)\Big| \, \mathrm d s \right)^2 \,\mathrm dt \right)^{1/2} \\
%
&= \left(\int_0^1 \left( M_1\int_0^1 \Big| f(s)\Big| \, \mathrm d s \right)^2 \,\mathrm dt \right)^{1/2} \\
%
&\leq \left(\int_0^1 \left( M_2\int_0^1 \Big| f(s)\Big| \, \mathrm d s \right)^2 \,\mathrm dt \right)^{1/2} \\
%
&= \left( \int_0^1 M_2^2 \left( \int_0^1 \Big| f(s)\Big| \, \mathrm d s \right)^2 \,\mathrm dt \right)^{1/2} \\
%
&= \big\Vert M_2 \Vert f\Vert_1^2\big\Vert_2 = M_2\big\Vert \Vert f\Vert_1^2\big\Vert_2\,.
\end{align}
This is great and all, but there is still that other pesky integral corresponding to $T$ inside the norm, which I want to get rid of. How could I achieve that? Should I use one of the big inequalitities (Hölder, Cauchy–Schwartz) to my benefit?
Best Answer
You seem to be missing an exponent in your $2$-norms. That being said, the easiest way to do this is probably just to use the Cauchy-Schwarz inequality pointwise to get that
$$ |Tf(t)|\leq \|f\|_2 \sqrt{ \int_0^1 e^{2(t+s^2)}\textrm{d}s} $$
Hence, $$ \| Tf(t)\|_2=\sqrt{\int_0^1 |Tf(t)|^2 \textrm{d}t}\leq \|f\|_2 \sqrt{\int_0^1 \int_0^1 e^{2(t+s^2)}}\textrm{d}s\textrm{d}t\leq \|f\|_2 e^2, $$ which proves that $T$ is bounded.