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While it is easy to see that $H^1(\mathbb{R})$ are Hölder $1/2$-continuous, I started wondering whether this implies that $\delta_x(\varphi)=\varphi(x)$ is continuous as a functional

$$\delta_x:H^1(\mathbb{R}) \rightarrow \mathbb{R}?$$

I believe it is false, but do not know a counterexample.

Since it was asked in the comments: Continuity means $$\left\lvert \varphi(x) \right\rvert \lesssim \left\lVert \varphi \right\rVert_{H^1}.$$

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    $\begingroup$ What would continuity mean in this case? Also, ask this at math.stackexchange.com. $\endgroup$
    – Dirk
    Commented Mar 23, 2018 at 7:55
  • $\begingroup$ @DIrk Well, $\left\lvert \varphi(x) \right\rvert \lesssim \left\lVert \varphi \right\rVert_{H^1}$ or course. $\endgroup$
    – Dixmier
    Commented Mar 23, 2018 at 7:56
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    $\begingroup$ As per comments below, could you please clarify what space $H^1$ is supposed to be - Sobolev, Hardy, ??? $\endgroup$
    – Nate Eldredge
    Commented Mar 23, 2018 at 13:25
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    $\begingroup$ At the level of abstract nonsense, once you know that all the elements of $H^1$ are continuous, the delta distribution basically has to be continuous. If $H^1$ functions are continuous (have continuous versions) then the inclusion map $H^1(\mathbb{R}) \to C([-1,1])$ is well defined. Morally, any everywhere-defined linear map between Banach spaces which didn't need the Axiom of Choice to construct, must be continuous. (More concretely, use the closed graph theorem to show continuity.) And the delta distribution is obviously continuous on $C([-1,1])$. $\endgroup$
    – Nate Eldredge
    Commented Mar 23, 2018 at 13:38
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    $\begingroup$ Though of course it is best to work explicitly and get the right constant if you can. But this shows in advance that it's guaranteed to work. $\endgroup$
    – Nate Eldredge
    Commented Mar 23, 2018 at 13:38

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Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$; see e.g. Wikipedia for the definition of $H^k$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H\"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} \begin{split} f(0)^2\le\int_0^1(b+|f(y)|)^2dy& \le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\\ &\le2(a+b)^2=2\|f\|_{H^1}^2, \end{split} \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Using here the interval $[0,a/b]$ instead of $[0,1]$, one can improve the above inequality to \begin{equation} |f(0)|\le2\sqrt{ab}=2\sqrt{\|f\|_2\,\|f'\|_2}\le \|f\|_{H^1}. \end{equation}

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  • $\begingroup$ Perhaps a reference for this norm of $H^1(\mathbb R)$ should be provided. $\endgroup$
    – Gerald Edgar
    Commented Mar 23, 2018 at 12:41
  • $\begingroup$ @GeraldEdgar : Done. $\endgroup$
    – Iosif Pinelis
    Commented Mar 23, 2018 at 12:44
  • $\begingroup$ OK we have a quandary here. I understood $H^1$ to be a Hardy space en.wikipedia.org/wiki/Hardy_space not a Sobolev space. I guess the OP has to tell us what he means: this is a classic case of "not clear what you are asking". $\endgroup$
    – Gerald Edgar
    Commented Mar 23, 2018 at 12:48
  • $\begingroup$ @GeraldEdgar : The OP wrote $H^1(\mathbb{R})$ rather than $H^1(\mathbb{T})$. $\endgroup$
    – Iosif Pinelis
    Commented Mar 23, 2018 at 12:51
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    $\begingroup$ Nitpick: assuming this is Sobolev, $\|f\|_{H^1}$ isn't $a+b$ but rather $\sqrt{a^2+b^2}$. The argument still works but some of the numerical constants may change. $\endgroup$
    – Nate Eldredge
    Commented Mar 23, 2018 at 13:31

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