I'm trying to solve this exercise.
I solved the points (a) and (b) but I'm not able to solve the point (c). I'm trying to proceed in this way:
By following the hint we have $uu''=uf$, then by using the relation $(uu')'=(u')^2+uu''$, we get $$\int_0^1(uu')' \, dx=\int_0^1(u')^2\, dx+\int_0^1uu'' \, dx.$$
Since $u(0)=u(1)=0$ and $u''=f$, we get $$\vert\vert u'\vert\vert_2^2=-\int_0^1uf \, dx.$$ Now I don't know how to proceed. Can you help me?
NB: Here $\vert\vert \bullet\vert\vert_p^p=\int_0^1(\bullet)^p$.
By the fundamental theorem of calculus we can write (here we use that $u(0)=0$) $$ u(x)=\int_0^x u'(t)\,dt. $$ Thus, by the triangle inequality, $$ \int_0^1 u(x)^2\,dx=\int_0^1\Bigl(\int_0^x u'(t)\,dt\Bigr)^2\,dx\leq \int_0^1\Bigl[\int |u'(t)|\,dt\Bigr]^2\,dx $$ Using Cauchy--Schwarz inequality on the inner integral, we get $$ \int_0^1 u(x)^2\,dx\leq \int_0^1\Bigl[\int_0^x 1^2\,dt\int_0^x |u'(t)|^2\,dt\Bigr]\,dx. $$ Calculating the first of the inner integrals and bounding the second one when we replace the bound $x$ by $1$, we find that $$ \|u\|_2^2=\int_0^1 u(x)^2\,dx\leq \int_0^1 x \,dx\cdot \int_0^1|u'(t)|^2\,dt=\frac{1}{2}\|u'\|_2^2.\tag{*} $$ Multiplying $u''=f$ with $-u$ and integrating by parts (remembering the Dirichlet conditions), and using Cauchy--Schwarz again, we find that $$ \|u'\|_2^2=\int_0^1 (u'(x))^2\,dx=\int_0^1 u''(-u)\,dx =\int_0^1 f(-u)\,dx\leq \|f\|_2\|u\|_2.\tag{**} $$ Combinging (*) and (**) we find the inequality we look for, $$ \|u'\|_2^2\leq \|f\|_2\|u\|_2\leq \|f\|_2\frac{1}{\sqrt{2}}\|u'\|_2. $$ Just divide by $\|u'\|_2$ and square.