What's the value of $\alpha$ satisfying $||f'||^2\ge \alpha||f||^2$? I am reading a paper about numerical analysis of a certain method for solving operator equation. Let our Hilbert space be $L^2[0,1]$, we define the subspace $D\in L^2[0,1]$ by
$$
D:=\{f\in C^2(0,1)\cap C^1[0,1] \ | f(0)=f(1)=0 \}.
$$
There is a line saying "It is well known that there exists an $\alpha>0$ such that the inequality
$$
\|f'\|^2\ge \alpha\|f\|^2
$$
holds uniformly for every $f\in D$." without giving a reference.
Since I am quite new to the field, I assumed that it is a common knowledge that such $\alpha$ exists. Could anyone please provide me with a reference for this fact?
My current knowledge includes basic functional analysis, mainly from Kreyszig's book, but very little of Sobolev space since I have just begun studying it. Any help would be very appreciated.
 A: If you just want the existence of some constant, then that's not so difficult. Getting the best constant takes work. Start with the fundamental theorem for $C^1$ functions, and apply Cauchy-Schwartz to $(f',1)$:
\begin{align}
           f(x) & = \int_{0}^{x}f'(t)dt \\
       |f(x)|^2 & \le \left(\int_{0}^{x}dt\right)\left(\int_{0}^{x}|f'(t)|^2dt\right)=x\int_{0}^{x}|f'(t)|^2dt.
\end{align}
Now integrate by parts
\begin{align}
    \int_{0}^{\pi}|f(x)|^2dx & \le \int_{0}^{\pi}x\int_{0}^{x}|f'(t)|^2dt dx \\
   & = \left.\frac{1}{2}(x^2-\pi^2)\int_{0}^{x}|f'(t)|^2dt\right|_{x=0}^{\pi}
   -\frac{1}{2}\int_{0}^{\pi}(x^2-\pi^2)|f'(x)|^2dx \\
   & = \frac{1}{2}\int_{0}^{\pi}(\pi^2-x^2)|f'(x)|^2dx \\
   & \le \frac{\pi^2}{2}\int_{0}^{\pi}|f'(x)|^2dx.
\end{align}
This is a common way to derive Sobolev types of inequalities. And once you have an inequality of this type for a dense subset such as $C^1[0,\pi]$, then the inequality extends to the full Sobolev space.
A: This is called Wirtinger's inequality ("second version" in Wikipedia's terminology): 
$$\pi^2 \int_0^1 f^2(x)\,dx \le \int_0^1 (f'(x))^2 \,dx$$ 
The constant is sharp, attained by $f(x)=\sin \pi x$. 
The article gives a proof of the first version. To prove the second version, apply the first one to the odd extension of $f$, namely 
$f(-x)= -f(x)$. This extension is still $C^1$ smooth, and has zero integral on the larger interval. 
