Bound on $f(0)^2$ by integrals of $f^2$ and $(f')^2$ on $[0,1]$. 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.
 A: 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.
A: One can do as follows. We may assume that $f(0)=1$. Let
$$
J(f) = \int_0^1 (f'(x)^2 + f(x)^2) dx \geq 0.
$$
We want to minimize $J(f)$ under the condition that $f(0)=1$. Let $F(f',f) = f'^2+f^2$. From variational calculus we know that any minimum satisfies
$$
F_f' - \frac{d}{dx} F_{f'}' = 0
$$
and $F_{f'}'(f'(1),f(1))=0$. This simplifies to solving
$$
f'' = f
$$
with the boundary conditions $f(0)=1$ and $f'(1)=0$. The solution is
$$
f(x) = -\tanh(1)\sinh(x) + \cosh(x).
$$
One then checks that for this $f$ we have that
$$
J(f) > \frac{1}{2} = \frac{1}{2} f(0)^2.
$$
This proves the inequality.
(In fact, we have $J(f) = \tanh 1 > 1/2$.)
A: I think the inequality is correct. Let
$$ f(x)=\sum_{n=0}^\infty(a_n\cos(n\pi x)+b_n\sin(n\pi x)). $$
This is because $\{\cos(n\pi x),\sin(n\pi x)\}_{n=0}^\infty$ is dense in $C^1((0,1))\cap C([0,1])$. Noting that, for integers $m, n$,
\begin{eqnarray}
&\int_0^1\cos^2(n\pi x)dx=\int_0^1\sin^2(n\pi x)dx=\frac12,\\
&\int_0^1\cos(n\pi x)\sin(m\pi x)dx=0,\\
&\int_0^1\cos(n\pi x)\cos(m\pi x)dx=0 \text{ if }m\neq n,\\
&\int_0^1\sin(n\pi x)\sin(m\pi x)dx=0 \text{ if }m\neq n,
\end{eqnarray}
we have
\begin{eqnarray}
\int_0^1(f(x))^2dx&=&\int_0^1(\sum_{n=0}^\infty(a_n\cos(n\pi x)+b_n\sin(n\pi x)))^2dx\\
&=&\int_0^1\sum_{m,n=0}^\infty(a_m\cos(m\pi x)+b_m\sin(m\pi x))(a_n\cos(n\pi x)+b_n\sin(n\pi x))dx\\
&=&\frac{1}{2}\sum_{n=0}^\infty (a_n^2+b_n^2),\\
\int_0^1(f'(x))^2dx&=&\int_0^1(\sum_{n=0}^\infty(-n\pi a_n\sin(n\pi x)+n\pi b_n\cos(n\pi x)))^2dx\\
&=&\int_0^1\sum_{m,n=0}^\infty mn\pi^2(-a_m\sin(m\pi x)+b_m\cos(m\pi x))(-a_n\sin(n\pi x)+b_n\cos(n\pi x))dx\\
&=&\frac{1}{2}\sum_{n=0}^\infty n^2\pi^2(a_n^2+b_n^2).
\end{eqnarray}
Thus
\begin{eqnarray}
&&\int_0^1(f(x))^2dx+\int_0^1(f'(x))^2dx-\frac12(f(0))^2\\
&=&\frac12\sum_{n=0}^\infty (n^2\pi^2+1)(a_n^2+b_n^2)-\frac12(\sum_{n=0}^\infty a_n)^2\\
&\ge&\frac12\left(\sum_{n=0}^\infty(n^2\pi^2+1)a_n^2-(\sum_{n=0}^\infty a_n)^2\right)+\frac12\sum_{n=0}^\infty(n^2\pi^2+1)b_n^2.
\end{eqnarray}
I believe that 
$$ \sum_{n=0}^\infty(n^2\pi^2+1)a_n^2-(\sum_{n=0}^\infty a_n)^2\ge 0$$
and now I do not have time to prove it. So
$$ \int_0^1(f(x))^2dx+\int_0^1(f'(x))^2dx\ge\frac12(f(0))^2. $$
