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 $$\left\lvert \varphi(x) \right\rvert \lesssim \left\lVert \varphi \right\rVert\_{H^1}.$$



Is the Delta distribution a continuous functional on $H^1(\mathbb{R})$?
Let $f:=\varphi$, $a:=\|f\|\_2$, $b:=\|f'\|\_2$, so that $\|f\|\_{H^1}=a+b$; see e.g. [Wikipedia](https://en.wikipedia.org/wiki/Sobolev_space) 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}


