How do you maximize the quotient ||f||/||f'|| of euclidean norms if f is to be a function on [0,1] which vanishes on the boundary?
$||g||^2 = \int_0^1g(x)^2\textrm{d}x$
I guess $f$ needs to be continuously differentiable for it to make sense or something.
Without loss of generality, assume $\|g\|=1$. Then $$ \begin{align} 0 &=\delta\int_0^1g(x)^2\,\mathrm{d}x\\ &=2\int_0^1g(x)\,\delta g(x)\,\mathrm{d}x\tag1 \end{align} $$ For each $\delta g(x)$ that satisfies $(1)$, we want $$ \begin{align} 0 &=\delta\int_0^1g'(x)^2\,\mathrm{d}x\\ &=2\int_0^1g'(x)\,\delta g'(x)\,\mathrm{d}x\\ &=-2\int_0^1g''(x)\,\delta g(x)\,\mathrm{d}x\tag2 \end{align} $$ To satisfy $(2)$ for all $\delta g(x)$ that satisfy $(1)$, we must have $$ g''(x)=\lambda g(x)\tag3 $$ for some constant $\lambda$.
Explanation of $\boldsymbol{(3)}$
Let $\lambda$ be so that $$ \int_0^1g(x)g''(x)\,\mathrm{d}x=\lambda\int_0^1g(x)^2\,\mathrm{d}x $$ Then, since $$ \int_0^1g(x)\overbrace{(g''(x)-\lambda g(x))}^{\delta g(x)}\,\mathrm{d}x=0 $$ our condition says that $$ \int_0^1g''(x)\overbrace{(g''(x)-\lambda g(x))}^{\delta g(x)}\,\mathrm{d}x=0 $$ Subtract $\lambda$ times the former from the latter to get $$ \int_0^1(g''(x)-\lambda g(x))^2\,\mathrm{d}x=0 $$ which implies that $g''(x)=\lambda g(x)$.