Let $\alpha(f)=|f(0)|^2+\int_0^1f(x)^2dx$,find the maximum value of $\alpha(fg)/\alpha(g)$ Let $\alpha(f)=|f(0)|^2+\int_0^1|f(x)|^2dx$ be a functional on $C^2[0,1]$, the space of real functions o $[0,1]$ whose 2-oder derivative is continuous. Determine $\textrm{sup}\{\alpha(fg)/\alpha(g): g\not=0\}$, where $f$ is fixed in $C^2[0,1]$.

If we consider $||f||:=\sqrt{\alpha(f)}$, then $||f||$ is indeed a norm on $C^2[0,1]$, so we may consider $||fg||/||g||$ in stead of $\alpha(fg)/\alpha(g)$. It seem that its sup is ||f||, but I still have some confusion.

The sumpremum is equal to $$\max\{|f(t)|^2:\ t\in[0,1]\},$$ i.e., the square of the infinity norm of $f$.
We need to maximize $\alpha(fg)$ for $g$ with $\alpha(g)=1$. So 
$$
\alpha(fg)=|f(0)|^2\,|g(0)|^2+\int_0^1|f(t)|^2\,(g(t)|^2\,dt\leq\|f\|_{\infty}^2.
$$
Now if, $\|f\|_\infty=|f(0)|$, choose a twice-differentiable function $g_n$ with $g_n(0)=\sqrt{1-\tfrac1n}$, $g_n(t)=0$ for all $t>\tfrac1n$, and $\int_0^1 g(t)^2\,dt=\sqrt{\tfrac1n}$. Then
\begin{align}
\alpha(fg_n)&=\left(1-\tfrac1n\right)|f(0)|^2+\int_0^1 |f(t)|^2\,|g(t)|^2\,dt\\[0.3cm]
&=\left(1-\tfrac1n\right)\|f\|_\infty^2+\int_0^{1/n} |f(t)|^2\,|g(t)|^2\,dt\\[0.3cm]
&\geq \left(1-\tfrac1n\right)\|f\|_\infty^2.
\end{align}
And if $\|f\|_\infty=|f(t)|$ for $t>0$, fix $\varepsilon>0$. By continuity there exists $\delta>0$ such that $|f(t)|\geq \|f\|_\infty\varepsilon$ on $(t-\delta,t+\delta)$. Now construct twice-differentiable $g$, supported on $(t-\delta,t+\delta)$, with $\int_0^1 |g(t)|^2\,dt =1$. Then
$$
\alpha(fg)=\int_{t-\delta}^{t+\delta}|f(t)|^2\,|g(t)|^2\,dt\geq(\|f\|_\infty^2-\varepsilon)\int_{t-\delta}^{t+\delta}|g(t)|^2\,dt=(\|f\|_\infty^2-\varepsilon).
$$
