2
$\begingroup$

What I want to prove is the following inequality

$\Vert u \Vert_{L^2[0,1]} \leq \Vert u' \Vert_{L^2[0,1]}, \quad u \in W^{1,2}_0 [0,1], $

which I am trying to derive from

$$ \int_\Omega \vert u(x+h) - u(x) \vert^p \leq \Vert \nabla u \Vert^p_{L^p(\Omega)} \Vert h \Vert_q^p, $$

where $\Omega\subset \mathbb{R}^n, 1\leq p < \infty, 1/p + 1/q = 1,$ and $u \in C_0^\infty (\Omega), h\in \mathbb{R}^n$ arbitrary.

Now, setting $h=1$ is probably the way to go, since then $u(t+h)$ vanishes, and we get the result directly. However, there are two problems.

  1. We have a closed interval $[0,1]$. Can we simply replace it with $(0,1)$? I just know that both intervals have the same Lebesgue measure, but until now I have always encountered oven intervals in Sobolev spaces.

  2. We have in general $W_0^{k,p}(\Omega) = \overline{C_0^\infty(\Omega)}$, where the closure is taken with respect to the norm of $W$. So $u$ is not necessarily in $C_0^\infty(\Omega)$, but there is a sequence which converges to it.

So basically I can take an $(u_n)_{n\in \mathbb{N}} \subset C_0^\infty(\Omega)$, such that $ u_n \rightarrow u $ in the W-norm, and I can show that the inequality holds for all $u_n$. But I am not sure about how to proceed.

Edit: $u_n \rightarrow u$ in the W-Norm implies $\Vert u_n \Vert_W \rightarrow \Vert u \Vert_W$, and if I am not mistaken we also have in this case $\Vert x\Vert_W = \Vert x' \Vert_{L^2}$ by the definition of the W-norm. This gives $\Vert u'_n \Vert_{L^2} \rightarrow \Vert u' \Vert_{L^2}$. So the only missing piece would be then to show that $\Vert u_n \Vert_{L_2} \rightarrow \Vert u \Vert_{L_2}$. Any ideas?

$\endgroup$
  • $\begingroup$ You should clarify which norm you mean by 'W-norm'. $\endgroup$ – daw Apr 6 '18 at 12:00
1
$\begingroup$

I think it is simpler to argue from scratch. Functions $u \in W^{1,2}_0 [0,1]$ are absolutely continuous on $[0,1]$ and satisfy $u(0) = u(1) = 0$. The fundamental theorem of calculus gives $$|u(x)| = |u(x) - u(0)| = \left| \int_0^x u'(t) \, dt \right| \le \sqrt x \|u'\|_{L^2[0,1]}$$ whenever $x \in [0,1]$. Thus $$ \|u\|_{L^2[0,1]}^2 = \int_0^1 |u(x)|^2 \, dx \le \|u'\|_{L^2[0,1]}^2 \int_0^1 x \, dx = \frac 12 \|u'\|_{L^2[0,1]}^2 .$$

$\endgroup$
  • $\begingroup$ Thanks for the comment. Though I am curios about whether my approach would also work, since I have come quite close (I think). $\endgroup$ – Mr. Realstone Apr 3 '18 at 16:22
0
$\begingroup$

To answer your concerns:

(1) It does not make a difference whether you use $[0,1]$ or $(0,1)$ in integrals. This distinction is important when you work with continuous functions.

I suppose you want to work with the norm $\|u\|_{H^1_0} = \|u'\|_{L^2}$ on $H^1_0$. Then things get more difficult than needed.

Now let $u_n\to u$ in $H^1_0$ with smooth $u_n$'s, i.e., $\|u_n'-u'\|_{L^2}\to0$. For such smooth functions the inequality was already established. This implies that $(u_n)$ is a Cauchy sequence in $L^2$, hence convergent to some limit $v\in L^2$. It remains to show $v=u$. It is easy to show that $u'=v'$ (pass to the limit in definition of weak derivative of $u_n$). Hence $u$ and $v$ only differ by a constant, which implies $u=v$.

If one starts with $H^1_0$ as the completion of $C_c^\infty$ with respect to the full $H^1$-norm, then the problem disappears: $u_n\to u$ in $H^1$ implies $u_n\to u$ and $u_n'\to u'$ in $L^2$, and you can pass to the limit in the inequality. Then one gets that $u\mapsto \|u'\|_{L^2}$ is an equivalent norm on $H^1_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.