Is $\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx $ controlled by $\|f\|_{L^p(0,1)}$ Let $f\in L^{p}(0,1)$ for some $1<p<\infty$.
Is $$\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx $$ controlled by $\|f\|_{L^p(0,1)}$  ?
Equivalently:
Does $\frac{\|f\|_{L^p(0,x)}}{x}$ behave like
$\frac{\|f\|_{L^p(0,1)}}{x^{\alpha}}$ for some $\alpha<1$
as $x\rightarrow 0^+$ ?
I think it is stupid to try to prove the pointwise inequality
$\|f\|_{L^p(0,x)}\leq C x^{1-\alpha}\|f\|_{L^p(0,1)}$. Simply because for each fixed "very small" $x$ we can find a function supported in $[0,x]$ so that $\|f\|_{L^p(0,x)}=\|f\|_{L^p(0,1)}$ and the inequality fails.

I think I figured this out.  A counterexample:
Let $f(x):=\frac{\chi_{[0,1]}(x)}{x^{\frac{1-\epsilon}{p}}}$. Then
$\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx=\frac{C}{\epsilon^{1+\frac{1}{p}}}$.
While $\|f\|_{L^p(0,1)}=\frac{C}{\epsilon^{\frac{1}{p}}}$.
