Brezis Probem 8.8 Let $u\in W^{1,p}(0,1)$ with $1<p<\infty$ then show that $u(0)=0$ iff $\frac{u(x)}{x}\in L^p(0,1)$. And moreover $$\big\Vert\frac{u(x)}{x}\big\Vert_{L^p(0,1)}\le \frac{p}{p-1}\Vert u^{'}\Vert_{L^p(0,1)}$$
and if $u\in W^{2,p}(0,1)$ with $1<p<\infty$ and $u(0)=0=u^{'}(0)$ then $\frac{u(x)}{x^2}$ and $\frac{u^{'}(x)}{x}$ is in $L^p(0,1)$ with $$\big\Vert\frac{u(x)}{x^2}\big\Vert_{L^p(0,1)}+\big\Vert\frac{u^{'}(x)}{x}\big\Vert_{L^p(0,1)}\le \frac{p}{p-1}\Vert u^{"}\Vert_{L^p(0,1)}$$
Any Help or suggestion on how to approach this problem.
 A: Write $$\frac{u(x)}{x} = \frac{u(0)}{x} + \frac{1}{x}\int_0^x u'(t)\, dt\tag{1}\label{eq}$$
Hardy's integral inequality yields
$$\left\|\frac{1}{x}\int_0^x u'(t)\, dt\right\|_{L_x^p(0,1)} \le \frac{p}{p-1}\|u'\|_{L^p(0,1)}\tag{2}\label{eq1}$$
Thus $\frac{u(x)}{x}\in L^p(0,1)$ if and only if $\frac{u(0)}{x}\in L^p(0,1)$, which is true if and only if $u(0) = 0$. Further, when $u(0) = 0$, \eqref{eq} and \eqref{eq1} yield the desired first estimate.
The second estimate should not have $p/(p-1)$ on the right-hand side, but some constant $C_p$ depending on $p$. Use the first part of the problem to show 
$$\left\|\frac{u'(x)}{x}\right\|_{L_x^p(0,1)}\le \frac{p}{p-1}\|u''\|_{L^p(0,1)}\tag{3}\label{eq2}$$
Now 
$$\left\lvert \frac{u(x)}{x^2}\right\rvert \le \frac{1}{x}\int_0^x \left\lvert \frac{u'(t)}{x}\right\rvert\, dt \le \frac{1}{x}\int_0^x \left\lvert \frac{u'(t)}{t}\right\rvert\, dt$$
so by Hardy's inequality and \eqref{eq2},
$$\left\|\frac{u(x)}{x^2}\right\|_{L_x^p(0,1)} \le \frac{p}{p-1}\left\|\frac{u(x)}{x}\right\|_{L^p(0,1)}\le \left(\frac{p}{p-1}\right)^2\|u''\|_{L^p(0,1)}\tag{4}\label{eq3}$$
Combining \eqref{eq2} and \eqref{eq3} results in
$$\left\|\frac{u(x)}{x^2}\right\|_{L_x^p(0,1)} + \left\|\frac{u'(x)}{x}\right\|_{L_x^p(0,1)} \le C_p\|u''\|_p$$
where $C_p = p/(p-1) + [p/(p-1)]^2$.    
