What's the value of $\alpha$ satisfying $||f'||^2\ge \alpha||f||^2$? [duplicate]

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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.

marked as duplicate by user147263, JMP, Shailesh, choco_addicted, user91500Mar 27 '16 at 8:56

• @404 Thank you a lot for your input! I didn't even know that it has that name, I'll carefully read the link you provide in your answer. Feel free to close it if you think it's a duplicate of an already existing topic. – BigbearZzz Mar 27 '16 at 5:58

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.
• @BigbearZzz : The best bound is found by expanding in a basis of eigenfunctions of $L=-\frac{d^2}{dx^2}$ with endpoint conditions $f(0)=f(1)=0$. The eigenfunctions are $\sin(\pi x),\sin(2\pi x),\sin(3\pi x),\cdots$. So the minimum bound is the minimum eigenvalue or, actually its square in your case because of how the squares appear. So the minimum bound is $\pi^2$. And that's because $(Lf,f)=\|f'\|^2$ for $f\in\mathcal{D}(L)$. The minimum bounds for Sobolev estimates of this type on more general regions come from the minimum eigenvalue of the Laplacian with Dirichlet conditions. – DisintegratingByParts Mar 27 '16 at 8:16
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.