Prove a function is in $L^2[0,1]$

Question

If $f\in L^2[0,1]$, and $$g(x)=\int_0^1\frac{f(t)\mathrm dt}{|x-t|^{1/2}},\quad x\in[0,1],$$ show that $\|g\|_2\le2\sqrt2\|f\|_2$.

I tried Minkowski's integral inequality (with $p=1/2$, so the inequality reverses), but cannot reach the inequality I need. I also used Holder's inequality and failed too.

What is the correct approach to solve this problem?

Answer 1

A possible solution steps:

1. Prove that $$ \int_0^1\frac{1}{|x-t|^{1/2}}\,dt=2\sqrt{x}+2\sqrt{1-x}\le 2\sqrt{2}. $$
2. Prove that $\|g\|_\infty\le 2\sqrt{2}\|f\|_\infty$ (simple estimation by 1).
3. Prove that $\|g\|_1\le 2\sqrt{2}\|f\|_1$ (using e.g. Tonelli's theorem and 1).
4. Conclude that $\|g\|_2\le 2\sqrt{2}\|f\|_2$ by the Riesz-Thorin theorem.

Answer 2

Here is another answer only uses basic calculus. From the expression of $g$, \begin{align} \|g\|_2^2 &=\int_0^1 g(x)^2 dx \cr & =\int_0^1\int_0^1 \int_0^1 \frac{f(t)}{|x-t|^{1/2}}\frac{f(s)}{|x-t|^{1/2}}dsdtdx \cr &\leq \int_0^1 \int_0^1 \int_0^2 \frac{f(t)^2+f(s)^2}{2}|x-t|^{-1/2}|x-s|^{-1/2}dsdtdx \cr &=\int_0^1 f(t)^2 \left[\int_0^1 |x-t|^{-1/2} \left(\int_0^1 |x-s|^{-1/2}ds\right)dx \right]dt. \end{align} Then we can get the desired inequality using the estimate $\int_0^1 |x-s|^{-1/2}ds=\int_0^1 |x-t|^{-1/2}dx\leq 2\sqrt{2}$.

Comment: The inequality is actually true for any $L^p$ norm using the interpolation theorem in the other answer.