Poincaré constant for subspace of Hilbert space

Question

Let $V=\{h \in H^1(0,L)\mid h(0)=0\}$. I wish to show a Poincaré-type inequality for $v\in V$ with a specific constant: $$\| v\|_{L^2(0,L)} \le \frac{2L}{\pi} \| v' \|_{L^2(0,L)}$$ I already have a sketch for a proof involving $L/\pi$ that uses Fourier series. I can't seem to get the better constant with the same approach.

Answer 1

I've found a way. First, consider the space $H_0^1(0,L)$. This is a subspace of $V$ and $H^1$. Let $h\in H_0^1(0,L)$, thus we can define: $$\tilde{h}=h \quad \text{for} \; x\ge 0, \quad -h \quad \text{for} \; x<0$$ $\tilde{h}$ is an odd, $2L$-periodic function. Its Fourier series will only contain odd terms: $$\tilde{h} = \sum_n b_n \sin(\pi nx/L)$$ And it's easy to see that this Fourier series converges uniformly to the function $\tilde{h}$, since the function is continuous on $\Bbb{R}$ and its derivative is piecewise continuous. Then $\tilde{h}'$ can be obtained by differentiating each term of the series: $$\tilde{h} = \sum_n \frac{\pi n b_n}{L} \cos(\pi nx/L)$$ Now applying Bessel's identity we get: $$||\tilde{h}||_{L^2(-L,L)}^2 = \sum_n b_n^2 \\\\ ||\tilde{h}'||_{L^2(-L,L)}^2 = \sum_n \Big(\frac{\pi n b_n}{L}\Big)^2 \\\\ \implies ||\tilde{h}'||_{L^2(-L,L)}^2 \ge \sum_n \Big(\frac{\pi b_n}{L}\Big)^2 = \frac{\pi^2}{L^2} \sum_n b_n^2 \\\\ \implies||\tilde{h}'||_{L^2(-L,L)} \ge \frac{\pi}{L} ||\tilde{h}||_{L^2(-L,L)}$$

We will now notice that: $$||\tilde{h}||_{L^2(-L,L)}^2 = 2||h||_{L^2(0,L)}^2 \\ ||\tilde{h}' ||_{L^2(-L,L)}^2 = 2||h'||_{L^2(0,L)}^2$$ And this allows us to say, finally, that: $$||h||_{L^2(0,L)} \le \frac{L}{\pi}||h'||_{L^2(0,L)}$$

To get the final inequality, note that for any function $v\in V$ we may define a function $\bar{v} \in H_0^1(0, 2L)$ as follows: $$\bar{v}(x) = v(x)\quad \text{for} \; 0\le x < L, \quad v(2L - x) \quad \text{for} \; L\le x \le 2L$$ Which is symmetric with respect to the vertical axis centered in $L$, thus $\bar{v} \in H_0^1(0, 2L)$. As before: $$||\bar{v}||_{L^2(0,2L)}^2 = 2||v||_{L^2(0,L)}^2 \\ ||\bar{v}' ||_{L^2(0,2L)}^2 = 2||v'||_{L^2(0,L)}^2$$ By the argument above, it immediately follows that: $$||v||_{L^2(0,L)} \le \frac{2L}{\pi}||v'||_{L^2(0,L)}$$