I am working on the following homework exercise:

Let $H^1_0([0,1]) := \overline{C^1_0([0,1])}$ w.r.t. $\|\varphi\|_{H^1_0} := (\int_0^1 |\varphi'(x)|^2 \mathrm{d}x)^{1/2}$.

  1. Let $\tilde{H}^1_0([0,1]) := \{\varphi \in H^1([0,1]) : \varphi(0) = \varphi(1) = 0\}$. Show that on $\tilde{H}^1_0([0,1])$ the norms $\|\cdot\|_{H^1_0}$ and $\|\cdot\|_{H^1}$ are equivalent.
  2. Show that $\|\cdot\|_{H^1_0}$ is not a norm on $H^1([0,1])$.
  3. Show that $\tilde{H}^1_0([0,1]) = H^1_0([0,1])$ (up to unitary equivalence).

Admittedly, I am quite lost. I know that $\|\varphi\|_{H^1} = (\int_0^1 |\varphi(x)|^2 + |\varphi'(x)|^2 \mathrm{d}x)^{1/2}$ and I am allowed to use the fact that $\|\varphi\|_\infty = \max |\varphi(x)| \leq C\|\varphi\|_{H^1}$. How can I apply this knowledge to question one? Any hints or starting points appreciated.

Edit 1: Thank you for the comments. I think I solved item one: $$ \|\varphi\|_{H^1_0} = \left(\int_0^1 |\varphi'(x)|^2 \mathrm{d}x\right)^{1/2} \leq \left(\int_0^1 \underbrace{|\varphi(x)|^2}_{\geq 0} + |\varphi'(x)|^2 \mathrm{d}x\right)^{1/2} = \|\varphi\|_{H^1} $$ and $$ \|\varphi\|_{H^1} = \left(\int_0^1 |\varphi(x)|^2 \mathrm{d}x + \int_0^1 |\varphi'(x)|^2 \mathrm{d}x\right)^{1/2}, $$ where \begin{align*} \int_0^1 |\varphi(x)|^2 \mathrm{d}x &= \int_0^1 \left|\int_0^x \varphi'(y)\,\mathrm{d}y\,\right|^2 \mathrm{d}x \\ &\leq \int_0^1\int_0^x |\varphi'(y)|^2 \mathrm{d}y\,\mathrm{d}x \\ &\leq \int_0^1\int_0^1 |\varphi'(y)|^2 \mathrm{d}y\,\mathrm{d}x \\ &= \int_0^1 |\varphi'(y)|^2 \mathrm{d}y. \end{align*} Hence $$ \|\varphi\|_{H^1} \leq \left(2\int_0^1 |\varphi'(x)|^2 \mathrm{d}x\right)^{1/2} = \sqrt{2}\|\varphi\|_{H^1_0}. $$

Am I correct with this?

I'm now thinking about the other two problems. Any ideas are welcome, of course. =)

Edit 2: For item two, consider $\psi(x) = 1, \psi \in H^1([0,1])$, then $\|\psi\|_{H^1_0} = 0$. Hence $\|\cdot\|_{H^1_0}$ fails definiteness on $H^1([0,1])$.

Edit 3: My attempt for problem three: $C^1_0$ is dense in $(H^1, \|\cdot\|_{H^1})$ because it is dense in $C^1$ and $C^1$ is dense in $H^1$. Hence, $C^1_0$ is dense in $(\tilde{H}^1_0, \|\cdot\|_{H^1})$ and $(\tilde{H}^1_0, \|\cdot\|_{H^1_0})$ because the norms are equivalent by item one. $\tilde{H}^1_0$ is a complete metric space because it is a closed subspace of $H^1$. Now, because the closure of a subspace of a complete metric space is its completion and because the completion of a metric space is unique, the two definitions $H^1_0$ and $\tilde{H}^1_0$ agree.

Could anybody confirm this reasoning (possibly also for the two previous problems)?

  • 2
    $\begingroup$ I feel like I had this as a homework exercise... maybe a Poincare type inequality? $\endgroup$ – Sean Roberson Jun 13 '17 at 17:00
  • $\begingroup$ Thank you. We did not cover Poincaré in the lecture however, so I'd prefer some other way. $\endgroup$ – bbrot Jun 13 '17 at 17:07
  • 1
    $\begingroup$ The norm equivalence in $1$ is basically a (special case of) Poincare. I think in one dimension if you start by replacing $\varphi$'s appearance in the $H^1$ norm by the integral of its derivative, you can see some inequalities that will get you the needed bound. $\endgroup$ – Jason Knapp Jun 13 '17 at 17:11
  • 1
    $\begingroup$ math.stackexchange.com/questions/1684452/… Seems like this would help with the first item. $\endgroup$ – Jason Knapp Jun 13 '17 at 17:28
  • $\begingroup$ What is your definition of $C_0^1$? I would guess you meant to write $C_c^1((0,1))$ instead of $C^1_0([0,1])$, otherwise there's not much to prove. Anyway, this shouldn't be dense in $C^1$. $\endgroup$ – Michał Miśkiewicz Jun 14 '17 at 16:31

Your solutions to 1 and 2 is correct. Michał Miśkiewicz is right, $C^1_0$ is not dense in $C^1$. Your solution to 3 is not correct. Given $\varphi\in \tilde{H}_0^1$ you need to construct a sequence $\varphi_n\in C^1_0$ which converges to $\varphi$ in $\tilde{H}_0^1$. First you should change $\varphi$ so that it is zero near $0$ and near $1$. Extend $\varphi$ to be zero outside of $(0,1)$. Since $\varphi(0)=\varphi(1)=0$ it follows that the extended function $\varphi\in H^1((-1,1))$ (you should prove this). Then define $\varphi_t(x)=\varphi(tx)$ for $t>1$. Note that $\varphi_t(x)=0$ for $x\in (1/t,1)$ and $\varphi_t\to \varphi$ as $t\to 1^+$ in $\tilde{H}_0^1$. Hence $\Vert \varphi-\varphi_t\Vert_{\tilde{H}_0^1}\le \varepsilon$ for $t$ close to $1$. Next define $g_{t,s}(x)=\varphi_t(x+s)$ for $s>0$ small. Then $g_{t,s}(x)=0$ for $x\in [0,s]$ and $g_{t,s}(x)=0$ for $x>\frac1t+s$. Since $\frac1t<1$ by taking $s$ small we have that $\frac1t+s<1$. Moreover, $g_{t,s}\to \varphi_t$ in $\tilde{H}_0^1$ as $s\to 0^+$. Hence $\Vert \varphi_t-g_{t,s}\Vert_{\tilde{H}_0^1}\le \varepsilon$ for $s$ small. Since $ g_{t,s}$ is zero in $[0,s]$ and in $ [\frac1t+s,1]$ you can now mollify it and you have that $\varphi_\epsilon*g_{t,s}$ is in $C^1_0$ and it converges to $ g_{t,s}$ as $\epsilon \to 0^+$. $\Vert \varphi_\epsilon*g_{t,s}-g_{t,s}\Vert_{\tilde{H}_0^1}\le \varepsilon$ for $\epsilon$ small

  • $\begingroup$ Sorry for the late reply and thank you very much. It seems I mixed things up. One more question: Is it true that $C^1_0$ is dense in $L^2$? $\endgroup$ – bbrot Jun 25 '17 at 21:17
  • $\begingroup$ yes, you use a similar proof of what I wrote above but it is even simpler since functions in $L^2$ can be discontinuous, so you first approximate your function with one which is zero near the boundary and then use mollifiers. $\endgroup$ – Gio67 Jun 26 '17 at 10:08

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