Finding infimum of a class of functions in $C[0,1]$ If $u$ is a continuously differentiable function in $[0,1]$ such that $u(0)=0$, max $|u|=1$ in $[0,1]$,then what is the infimum of $\int_0^1(u'(t))^2 dt$?
 A: Let $x_0\in (0,1]$ be a maximum point for $|u|$, i.e. $|u(x_0)| = 1$.
Then, by Holder's inequality,
$$
(*) \qquad 1 = |u(x_0)| \leq \int_0^{x_0} |u'| \leq \sqrt{x_0} \left(\int_0^{x_0}|u'|^2\right)^{1/2}.
$$
Hence [simplified version suggested by zhw]
$$
F(u) := \int_0^1 |u'|^2 \geq \int_0^{x_0}|u'|^2 \geq \frac{1}{x_0}\,.
$$
Now it is easily seen that the minimum of $F$ is achieved for $x_0=1$ by the function $\overline{u}(x) = x$.
[Long version, maybe it clarifies the conclusion above.]
It is easily seen that the function 
$$
v(x) := \begin{cases}
u(x), & \text{if}\ x \in [0,x_0],\\
\text{sign} u(x_0), & \text{if}\ x \in [x_0, 1],
\end{cases}
$$
satisfies $F(v) := \int_0^1 |v'|^2 \leq \int_0^1 |u'|^2$ (with strict inequality if $x_0<1$ and $u$ is not constant in $[x_0, 1]$), hence it is not restrictive to consider, in our minimization problem, only those functions which are constant after $x_0 \in \text{argmax} |u|$.
On the other hand, among all competing functions of this kind with maximum of $|u|$ attained at $x_0\in (0,1]$, we have that
$$
\overline{u}(x) :=
\begin{cases}
x/x_0, & \text{if}\ x\in [0, x_0],\\
1, & \text{if}\ x\in [x_0, 1],
\end{cases}
$$
attains the minimum of $F$, since
$$
(**) \qquad
F(\overline{u}) = \int_0^{x_0} \frac{1}{x_0^2}\, dx = \frac{1}{x_0}
$$
whereas, by (*),
$$
F(u) \geq \frac{1}{x_0} 
$$
with strict inequality if $u'$ is not constant in $[0, x_0]$.
Finally, by (**), the functional is minimized when $x_0 = 1$.
