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I don't understand an estimate in my textbook, maybe you can help me out!

Lemma (Poincaré's inequality).

Let $Ω ⊂ (0,L) × \mathbb R^{n−1}$. For $u ∈ C^\infty_c(Ω)$ we have the estimate $$ \int_Ω |u|^2\, dx ≤ L^2 \int_Ω |∇u|^2 \, dx.$$

Proof.

Extend $u$ with $u(x) = 0$ for $x \not \in Ω$. For $x = (x^1,x′) ∈ Ω$ estimate $$ |u(x^1,x')|^2 = \bigg|\int_0^{x_1} \frac{\partial u}{\partial x_1} (s,x') \, ds \bigg|^2 \leq \bigg(\int_0^L \bigg|\frac{\partial u}{\partial x_1}(s,x')\bigg| \, ds \bigg)^2 \leq L \int_0^L |\nabla u(s,x')|^2 \, ds.$$

Then it follows

$$ \int_\Omega |u|^2 \, dx \leq \int_{\mathbb R^{n-1}} \int_0^L |u(x_1,x')|^2 \, dx_1 \, dx' \leq \int_{\mathbb R^{n-1}} L^2 \int_0^L |\nabla u(s,x')|^2 \, ds \, dx' \leq L^2 \int_\Omega |\nabla u|^2 \, dx.$$ $\Box$

I very well understand the first inequalities, via Hölder's inequality.

But in the second line i don't understand where the second $L$ comes from.

And shouldn't the last inequality be an equality?

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  • $\begingroup$ How did you get the inequality $\left( \int_{0}^{L} \left| \frac{\partial u}{\partial x_1} (s,x') \right| \,ds \right)^2 \leq L \int_{0}^{L} |\nabla u(s,x') |^2 \,ds$ ? $\endgroup$
    – Leonidas
    May 20, 2023 at 18:04

1 Answer 1

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In the first inequality, integrate with respect to $x_1$ from $0$ to $L$. Since the right hand side is independent of $x_1$ you end up with $$\int_0^L \lvert u(x_1,x')\rvert^2 dx_1 \le L^2 \int^L_0 \lvert \nabla u(s,x') \rvert^2 ds.$$ This is the inequality you apply to derive the second one.

EDIT: The initial inequality that proved is $$\lvert u(x_1,x') \rvert^2 \le L \int^L_0 \lvert \nabla u(s, x') \rvert^2 ds.$$ In this inequality, the left hand side depends on $x_1$, but the right hand side does not. Thus we can integrate with respect to $x_1$ from $0$ to $L$ and switch the order of integration on the right (I suppose this requires Tonelli's theorem): \begin{equation} \begin{aligned} \int^L_0 \lvert u(x_1,x')\rvert^2 dx_1 \le L \int^L_0 \int^L_0 \lvert \nabla u(s,x') \rvert^2 dsdx_1 &= L \int^L_0 \lvert \nabla u(s,x') \rvert^2 \left( \int^L_0 dx_1 \right) ds \\ &=L^2\int^L_0 \lvert \nabla u(s,x') \rvert^2 ds. \end{aligned} \end{equation} Then we see \begin{align*} \int_\Omega |u|^2 \, dx &= \int_{\mathbb R^{n-1}} \left[\int_0^L |u(x_1,x')|^2 \, dx_1\right] \, dx' \\& \leq \int_{\mathbb R^{n-1}} \left[L^2 \int_0^L |\nabla u(s,x')|^2 \, ds\right] \, dx' \\ &= L^2 \int_\Omega |\nabla u|^2 \, dx.\end{align*} where the inequality of the two terms in the square braces is exactly the inequality we arrived at above.

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    $\begingroup$ I'm sorry, i don't understand your argument. could you elaborate a bit? $\endgroup$ Mar 10, 2017 at 23:12
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    $\begingroup$ @augustinsouchy I added details $\endgroup$
    – User8128
    Mar 11, 2017 at 0:58
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    $\begingroup$ awesome, thanks! (i think you're missing the squares on the first 8 integrals!) $\endgroup$ Mar 11, 2017 at 8:36
  • $\begingroup$ Ah, of course I do. Edited now. Sorry about that $\endgroup$
    – User8128
    Mar 11, 2017 at 18:03

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