Is $\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx $ controlled by $\|f\|_{L^p(0,1)}$

Question

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.

Answer 1

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 $\|f\|_{L^p(0,1)}=\frac{C}{\epsilon^{\frac{1}{p}}}$.