A linear operator on the complex Lebesgue space $L^2$ Let $X$ be the complex Lebesgue space $L^2(0,1)$. Let $T:X\to X$ be
$(Tf)(x)=x\int_0^1 \int_0^r f(s)\ ds\ dr-\int_0^x\int_0^r f(s)\ ds\ dr$
Show that $|Tf(x)|\leq \lVert f \rVert$ for any $0\leq x \leq 1$ and $f\in X$.
By writing $|Tf(x)|\leq |\int_0^1 \int_0^r f(s)\ ds \ dr|+|\int_0^x \int_0^r f(s) \ ds \ dr| \leq 2\int_0^1 \int_0^1 |f(s)| \ ds\ dr=2 \lVert f \rVert_{L^1}$, I can bound $|Tf(x)|$ by the $L^1$-norm of $f$. However the $L^2$-norm is less than the $L^1$-norm. I may need to find a way to bound it tighter.

We can write
$$Tf(x) = (x-1)\int_0^x \int_0^r f(s)\,ds\,dr + x\int_x^1\int_0^r f(s)\,ds\,dr.$$
To estimate each part, we use the Cauchy-Schwarz inequality, giving
$$\Biggl\lvert \int_0^r1 \cdot f(s)\,ds\Biggr\rvert \leqslant \sqrt{r}\cdot \lVert f\rVert,$$
and hence
\begin{align}
\Biggl\lvert(x-1)\int_0^x\int_0^r f(s)\,ds\,dr\Biggr\rvert
&\leqslant \lvert x-1\rvert\cdot\lVert f\rVert\int_0^x \sqrt{r}\,dr \\
&= \frac{2}{3}x^{3/2}(1-x)\lVert f\rVert
\end{align}
and
\begin{align}
\Biggl\lvert x\int_x^1\int_0^r f(s)\,ds\,dr\Biggr\rvert
&\leqslant x\cdot \lVert f\rVert \int_x^1 \sqrt{r}\,dr \\
&= \frac{2}{3}\bigl(1 - x^{3/2}\bigr)x\lVert f\rVert.
\end{align}
It remains to see
$$x^{3/2}(1-x) + \bigl(1-x^{3/2}\bigr)x \leqslant \frac{3}{2}.\tag{$\ast$}$$
Each of the two terms is clearly $\leqslant 1$. If $x \in [0,1/2]$, then the second term is $\leqslant \frac{1}{2}$ since  $1 - x^{3/2} \leqslant 1$, and if $x \in [1/2,1]$, then the first term is $\leqslant \frac{1}{2}$. So $(\ast)$ holds.
