Assume $y \in C^2[a,b]$ with $y(a)=y(b)=0$. Also assume, $$\|y\|+\|y'\| \leq 1$$ where $\|\cdot\|$ denotes the supremum norm on $C[a,b]$. I want to show that there exists a (uniform on all such $y$) positive constant $c$ such that $$\int\limits_a^b \left(2|y(t)y'(t)| + (y(t))^2\right)dt \leq c\int\limits_a^b(y'(t))^2dt.$$
The book I'm reading says to use the Cauchy-Schwarz inequality. But applying it to the left hand side blindly doesn't seem to give anything. I tried applying it pointwise:
$$\int \left(2|y|,y\right)\cdot\left(|y'|,y\right)dt \leq \left(\int \sqrt{4y^2+y^2}\right)\left(\int\sqrt{(y')^2+y^2}\right).$$
And also on the inner product on $C[a,b]$:
$$\int 2|yy'|+\int y^2 \leq 2\left(\int y^2\right)^{1/2}\left(\int(y')^2\right)^{1/2} + \int y^2.$$
What am I missing? Why would one even believe such a statement? Please send help.
Since $\|y\| \leq 1$ you have no problem with the first term. Your problem is with $\int y^{2}(t)\, dt$. To hang=dle this use the fact that $|y(t)|=|\int_a^{t} y'(s)\, ds| \leq \sqrt {\int_a^{b} y'(s)^{2}\,ds} \sqrt {b-a}$. Can you complete the argument now?