An inequality using Cauchy inequality Let $f\in L^2([0,1])$ and $$g = \int_0^1 \frac{f(t)}{|x-t|^{0.5}}dt.\quad 0<x<1.$$ Prove the following inequality holds:
$$||g||_2 \le 2\sqrt{2}||f||_2.$$
I want to use the fact that $||g|| = max ||gh||_1$ for  $h\in L^2([a,b])$ with $||h||_2=1.$ For arbitrary $h\in L^2$ we have $|gh|_1 = \int_0^1|h(x)|dx(|\int_{-1}^1 \frac{f(x-t)}{|t|^{0.5}}\kappa_{0\le x-t\le 1}(t)dt|)\le \int_{-1}^1\frac{1}{|t|^{0.5}}dt(\int_0^1|h(x)||\tilde{f}(x-t)|dt) \le \int_0^1\frac{1}{|t|^{0.5}}||f||_2||h||_2dt = 2||f||_2$ by Tonelli theorem, here $\tilde{f} = f\kappa_{0\le x-t\le 1}(t).$ Thus I found the coefficient can be greater than $2\sqrt{2}$. But I couldn't find the error in my proof. Could anyone help?
 A: $\def\abajo{\\[0.3cm]}$
Your estimate is wrong since with $f(t)=1$ you get $\|g\|_2=\sqrt{4+\pi}>2$. I don't follow exactly what you did, but it looks like you are missing a term. Below is my take, which I think is very similar to your intention, but using the missing term.
Assume without loss of generality that $f=|f|$, $h=|h|$. Write $a_v=1_{[v,1]}$, $b_v=1_{[0,1-v]}$
\begin{align}
\int_0^1|g(x)h(x)|\,dx
&\leq \int_0^1\int_0^1\frac{h(x)f(t)}{|x-t|^{1/2}}\,dt\,dx\abajo
&\textit{(substitution: $v=t-x$)}\abajo
&= \int_0^1\int_{-x}^{1-x}\frac{h(x)f(v+x)}{|v|^{1/2}}\,dt\,dx\abajo
&\textit{(Tonelli; substitution is explained at the end)}\abajo
&=\int_{-1}^0\frac1{(-v)^{1/2}}\int_{-v}^1h(x)f(v+x)\,dx\,dv
+\int_{0}^1\frac1{v^{1/2}}\int_{0}^{1-v}h(x)f(v+x)\,dx\,dv\abajo
&=\int_{0}^1\frac1{v^{1/2}}\bigg[\int_{v}^1h(x)f(x-v)\,dx\,dv
+\int_{0}^{1-v}h(x)f(v+x)\,dx\bigg]\,dv\abajo
&\textit{(Cauchy-Schwarz twice on each term, also explained below)}\abajo
&\leq\int_{0}^1\frac1{v^{1/2}}\Big[\|h\|_2\,\|f\|_2\,(\|a_v\|_2^2+\|b_v\|_2^2)\Big]\,dv\abajo
&=\|f\|_2\,\int_0^1 v^{-1/2}\,2(1-v)\,dv=\frac83\,\|f\|_2.
\end{align}
The constant $8/3$ is slighly less than $2\sqrt2$, so I assume that whoever asked the question had a different, more sleek argument in mind.

The substitution: The region of integration after the first substitution is, with $x$ as the horizontal axis and $v$ as the vertical one,

To exchange $x$ and $v$ requires two different sets of inequalities, which is what gives the two integrals.

Cauchy Schwarz: what's happening is
\begin{align}
\int_v^1 h(x)\,f(v+x)\,dx
&=\int_0^1 h(x)\,a(x)\,f(x+v)\,a(x)\,dx\abajo
&\leq \|h\|_2\,\|a\|_2\,\|f\|_2\,\|a\|_2.
\end{align}
The other integral is the same, but using the function $b$.
