Consider the class of functions


$$X:=\{f\in \mathcal{C}\_0^{\infty}(\mathbb{R})\;s.t.\;f\equiv 1 \mbox{ in a neighbourhood of}\;\;x=0\}$$


**Is it true that, for every $\varepsilon > 0$, I can find $f\in X$ such that $\|f\|\_{H^{1/2}(\mathbb{R})}<\varepsilon$?**


For $s\in[0,1/2)$, it is easy to show that the analogous question has affirmative answer. Indeed, given $f\in X$ and $n>0$, define $f\_n(x):=f(nx)$. Then $f\_n\in X$, and $\|f\_n\|\_{H^{s}(\mathbb{R})}\to 0$ as $n\to +\infty$.


Instead, when $s>1/2$, the analogous question has negative answer. Indeed, by Sobolev embedding one has
$$\|f\|\_{H^s}\geqslant C\|f\|\_{L^{\infty}}\geqslant C|f(0)|= C$$
for any $f\in X$.


In the critical case $s=1/2$ I think the answer is affirmative, but I'm not able to prove it.
Thank you for any suggestion.



Smallness of cut-off functions at critical Sobolev regularity
It is well known and easy to verify that (Exercise 14 p. 309 in [1])
$$
\log\Big|\log\sqrt{x^2+y^2}\Big|\in H^1(B^2(0,e^{-1}))
$$
so the trace of this function on the $x$-axis belongs to the trace space
$$
f(x)=\log\Big|\log|x|\Big|\in H^{1/2}((-e^{-1},e^{-1})).
$$
Let
$$
f\_t(x)=\begin{cases} 0 & \text{if } f(x)\leq t\\
f(x)-t & \text{if } t\leq f(x)\leq 2t\\
t & \text{if } f(x)\geq 2t
\end{cases}
$$
be a truncation of the function $f$ between the levels $t$ and $2t$, $t>0$. Then
$f\_t\in H^{1,2}$ and $\Vert f\_t\Vert\_{1/2}\leq\Vert f\Vert\_{1/2}$. 
Indeed, the space $H^{1/2}(\mathbb{R})$ is equipped with the norm
$$
\Vert u\Vert\_{1/2}=
\Vert u\Vert\_2+
\left(\int\_{\mathbb{R}}\int\_{\mathbb{R}}\frac{|u(x)-u(y)|^2}{|x-y|^2}\, dx\, dy\right)^{1/2}.
$$
Since $|f\_t|\leq|f|$ and $|f\_t(x)-f\_t(y)|\leq |f(x)-f(y)|$, it immediately follows that $\Vert f\_t\Vert\_{1/2}\leq\Vert f\Vert\_{1/2}$. Therefore
$$
\left\Vert\frac{1}{t}f\_t\right\Vert\_{1/2}\leq \frac{1}{t}\Vert f\Vert\_{1/2}\to 0
\quad
\text{as $t\to\infty$.}
$$
The function $t^{-1}f\_t$ equals $1$ near $0$ and it has compact support so approximating this function by convolution we can obtain a function $g\_t\in C\_0^\infty(\mathbb{R})$ such that $g\_t=1$ near $0$ and $\Vert g\_t\Vert\_{1/2}<\varepsilon$, provided $t$ is sufficiently large.


[1] **L. C. Evans,** *Partial differential equations.* Second edition. Graduate Studies in Mathematics, 19. American Mathematical Society, Providence, RI, 2010.


