# Obtaining a bound on the differential operator

I just need a little bit of help filling in the missing details in the following passage from Reddy (1986)'s Applied Functional Analysis and Variational Methods in Engineering

Let $C_0[0,1]$ be the space of continuous functions $u$ on [0,1] such that $u(0)=u(1)=0$. The differential operator $D\equiv \frac{d}{dx}$ is bounded below with respect to the $L_2$ norm in $C_0[0,1]$. We have $$u(x)=\int_0^x \frac{du}{dy} dy \leq \left[ \int_0^x \left|\frac{du}{dy}\right|^2 dy \right]^{\frac{1}{2}}$$

$$\int_0^1|u(x)|^2dx\leq\int_0^1 \left| \frac{du}{dx}\right|^2 dx.$$

I'm not quite sure what property allows me to conclude the first inequality. It almost looks like an application of Minkowski's inequality except for the fact that $\frac{du}{dy}\neq \left[ \left(\frac{du}{dy} \right)^2\right]^\frac{1}{2}$ (at least, not necessarily). I'm also not sure how the second inequality was obtained from the first.

Any help filling in the gaps would be greatly appreciated.

The first seems to be an application of Cauchy-Schwarz but I'm not $100\%$ sure on that. – Cameron Williams Jul 31 '13 at 1:34
It is Holder (use Holder inequality with the functions $\frac{du}{dy}$ and the constant function $1$) combined with the fact that $x\leq 1$: $$\int_0^x \frac{du}{dy}dy\leq \int_0^x \left|\frac{du}{dy}\right|dy\leq \left(\int_0^x \left|\frac{du}{dy}\right|^2dy\right)^{1/2}x^{1/2}$$
This almost makes sense. But the left hand side of the inequality uses $\frac{du}{dy}$, not $\left| \frac{du}{dy}\right|$. Unless it is somehow implicitly assumed that $\frac{du}{dy}>0$, I just can't see how we can conclude this. – Paul Jul 31 '13 at 1:42