Two questions about a function in $W^{1, p}(0, 1)$.

Question

1. Let $f \in W^{1, p}(0, 1)$ with $1 < p < \infty$. If $f(0) = 0$, then does it necessarily follow that$${{f(x)}\over x} \in L^p(0, 1)$$and$$\left\|{{f(x)}\over x}\right\|_{L^p(0, 1)} \le {p\over{p - 1}}\|f'\|_{L^p(0, 1)}?$$
2. Conversely, assume that $f \in W^{1, p}(0, 1)$ with $1 \le p < \infty$ and that$${{f(x)}\over x} \in L^p(0, 1).$$Does it necessarily follow that $f(0) = 0$?

Answer 1

Because $f$ is absolutely continuous, we have that $$f(x)=\int_0^x f'(t)dt.$$

Therefore $$\int_0^1 \frac{|f(x)|^p}{x^p}=\int_0^1 \left(\frac{1}{x}\left|\int_0^x f'(t)dt\right|\right)^p,$$

which implies, by using Hardy's inequality (see Sally's comment above and the link here) that $$\int _0^1 \left(\frac{|f(x)|}{x}\right)^p\le \left(\frac{p}{p-1}\right)^p\int _0^1 |f'(x)|^p,$$

whence $$\left\|\frac{f(x)}{x}\right\|_p\le \frac{p}{p-1}\|f'\|_p.$$

On the other hand, by using the fact that Hardy's inequality holds if and only if, the function being integrated is zero (see here again), we conclude that $$f'=0,\ a.e.,$$

however, if $f'$ is equal to zero, the right hand side is zero, so the left hand side must also be zero, which will be true only if $f=0$.