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?

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

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 $$ \|u\|_{L^2[0,1]}^2 = \int_0^1 |u(x)|^2 \, dx \le \|u'\|_{L^2[0,1]}^2 \int_0^1 x \, dx = \frac 12 \|u'\|_{L^2[0,1]}^2 .$$

  • $\begingroup$ 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

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 $H^1_0$.


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