TITLE: Bound on $f(0)^2$ by integrals of $f^2$ and $(f')^2$ on $[0,1]$.
QUESTION [6 upvotes]: Let $f$ be a function which is $C^1((0,1))\cap C([0,1])$. I would like to be able to show
$$
 \frac{1}{2}f(0)^2 \leq \int_0^1 f(x)^2dx + \int_0^1f'(x)^2dx
$$
where we are assuming that $f$ is a real-valued function.
We have attempted to use Young's inequality to reduce to $\frac{d}{dx}[f(x)^2]$, but this does not work.
Thanks in advance for any ideas!
edit: Counterexamples are welcome, obviously, but we believe this at the moment.

REPLY [3 votes]: Here is a variant which will yield an improved, but not optimal, constant in the inequality.
Let $\lambda$ be a positive real number whose exact value will be specified later.
Application of the fundamental theorem of calculus and the arithmetic-geometric inequality gives
\begin{eqnarray}
f(0)^2 &=& f(x)^2 - \int_0^x 2f(\xi)f'(\xi)\,d\xi \nonumber\\
&\leq& f(x)^2 + \int_0^x \frac{1}{\lambda}f(\xi)^2 + \lambda f'(\xi)^2\, d\xi \nonumber\\
&\leq& f(x)^2 + \int_0^1 \frac{1}{\lambda}f(\xi)^2 + \lambda f'(\xi)^2\, d\xi.
\end{eqnarray}
If we integrate both sides of this inequality from $0$ to $1$ and, moreover, choose
$\lambda$ such that $1+1/\lambda=\lambda$, that is $\lambda = (1+\sqrt{5})/2$, then we arrive at
\begin{equation}
\frac{2}{1+\sqrt{5}}f(0)^2 \leq \int_0^1 f(x)^2 + f'(x)^2 \, dx.
\end{equation}
As $2/(1+\sqrt{5}) \approx 0,6180 > 1/2$, the desired inequality follows.