Inequality $\int^{1}_{0}(u(x))^2\,\mathrm{d}x \leq \frac{1}{6}\int^{1}_{0} (u'(x))^2\,\mathrm{d}x+\left(\int_{0}^{1} u(x)\,\mathrm{d}x \right)^2$ I've been scratching my brain on this one for about a week now, and still don't really have a clue how to approach it.
Show that for $u \in C^1[0, 1]$ the following inequality is valid:
$$\int^{1}_{0}(u(x))^2\,\mathrm{d}x \leq \frac{1}{6}\int^{1}_{0} (u'(x))^2\,\mathrm{d}x+\left(\int_{0}^{1} u(x)\,\mathrm{d}x \right)^2$$
 A: Particular Case.  Suppose that $\int_0^1u(x)dx=0$. 
In this case we have
$$
u(x)=C+\int_0^xu'(t)dt 
$$
and $C$ is defined by the condition $\int_0^1u(x)dx=0$, this yields
$$\eqalign{
0&=C+\int_0^1\left(\int_0^xu'(t)dt\right)dx\cr
&=C+\left[(x-1)\int_0^xu'(t)dt\right]_0^1+\int_0^1(1-x)u'(x)dx
}
$$
Thus
$$\eqalign{
u(x)&=\int_0^xu'(t)dt-\int_0^1(1-t)u'(t)dt\cr
&=\int_0^1K(x,t)u'(t)dt
}
$$
where
$$
K(x,t)=\cases{t&if $t<x$,\cr t-1&if $t\geq x$.}
$$
So, by Cauchy-Schwarz
$$
(u(x))^2\leq \int_0^1(K(x,t))^2dt \int_0^1(u'(t))^2dt
$$
and consequently,
$$
\int_0^1(u(x))^2\leq \int_0^1\left(\int_0^1(K(x,t))^2dt\right)dx \int_0^1(u'(t))^2dt
$$
But it is easy to see that
$$\int_0^1\left(\int_0^1(K(x,t))^2dt\right)dx =
\int_0^1\left(\frac{x^3}{3}-\frac{(x-1)^3}{3}\right)dx=\frac{1}{6}
$$
So, 
$$
\int_0^1(u(x))^2\leq \frac{1}{6} \int_0^1(u'(t))^2dt
$$
Which is the desired inequality in this particular case.
The General case Case. We just apply the particular case to the function
$$
x\mapsto u(x)-\int_0^1u(t)dt
$$
and we note that
$$
\int_0^1\left(u(x)-\int_0^1u(t)dt\right)^2dx=
\int_0^1\left(u(x)\right)^2dx-\left(\int_0^1u(t)dt\right)^2.
$$
The general inequality follows.$\qquad\square$
A: By approximating the $u(x)$ through a linear combination of $S_n(x)=\sin(2\pi n x)$, 
$$ u(x) = \sum_{n\geq 0}a_n\,S_n(x)$$
we have:
$$\int_{0}^{1} S_n(x)\,S_m(x)\,dx = \frac{\delta_{m,n}}{2},\qquad\int_{0}^{1}u(x)^2\,dx = a_0^2+\frac{1}{2}\sum_{n\geq 1}a_n^2,\qquad\int_{0}^{1}u(x)\,dx = a_0\tag{1}$$
and:
$$u'(x) = \sum_{n\geq 1}2\pi\, n\,a_nC_n(x),$$
where $C_n(x)=\cos(2\pi n x)$. Since:
$$\int_{0}^{1}(u'(x))^2\,dx = 2\pi^2\sum_{n\geq 1}n^2 a_n^2$$
we just need to prove:
$$\sum_{n\geq 1}a_n^2\leq 4\zeta(2)\sum_{n\geq 1}n^2 a_n^2.$$
that is trivial. Wirtinger's inequality (more or less, we have just followed the lines of its proof) gives that the constant $\frac{1}{6}$ can be replaced by $\frac{1}{\pi^2}$.
We have assumed $f\in H^1([0,1])$. Obviously, if the weak derivative of $u(x)$ does not belong to $L^2([0,1])$, the given inequality is meaningless.
A: The constant $\frac{1}{6}$ can be slightly improved without using the full strength of Wirtinger Inequality.
$$\begin{align} \int_0^1 u^2(t)\,dt - \left(\int_0^1 u(t)\,dt\right)^2 & = \frac{1}{2}\int_0^1\int_0^1 \left(u(x) - u(y)\right)^2\,dx\,dy \\ &= \int_0^1 \int_x^1 \left(\int_x^y u'(t)\,dt\right)^2 \,dy\,dx \\ & \le \int_0^1\int_x^1 (y-x)\left(\int_x^y (u'(t))^2\,dt\right) \,dy\,dx \tag{1} \\ & = \int_0^1 (u'(t))^2 \int_0^t \int_t^1 (y-x)\,dy\,dx\,dt \tag{2} \\ & = \frac{1}{2}\int_0^1 t(1-t)(u'(t)) ^2 \,dt \\ & \le \frac{1}{8} \int_0^1 (u'(t))^2\,dt \tag{3}\end{align}$$
$(1)$ Cauchy Schwarz Inequality.
$(2)$ Change in order of integration.
$(3)$ $\displaystyle t(1-t) \le \frac{1}{4} \textrm{ for } t \in (0,1)$.
