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.

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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 $x\leq 1$: $$\int_0^x \frac{du}{dy}dy\leq \int_0^x \left|\frac{du}{dy}\right|dy\leq \left(\int_0^x \left|\frac{du}{dy}\right|^2dy\right)^{1/2}x^{1/2}$$

Remark: This inequality is also known as Poincaré inequality.

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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