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.

I now think this is indeed possible, and here's a sketch of my current ideas: I want to use the formula $$ \|f\|_{H^{1/2}}^2 \simeq \|f\|_2^2 + \int\!\!\int \left( \frac{f(y)-f(x)}{y-x} \right)^2 \, dxdy $$ to compute the $H^{1/2}$ norms. I'll focus on our function on $x<0$, and I'll let it increase from $0$ to $1$ on $[-\epsilon, 0]$ (in general, it seems clear that passing to the increasing rearrangement will only decrease the $H^{1/2}$ norm).

If we just use a linear function, then we obtain a contribution of $1$ from the double integral. Now split this up (in Cantor function style) and do the increase on two tiny intervals $I_j$ of size $\delta\ll\epsilon$ each, and $f=1/2$ on an interval in the middle of length almost $\epsilon$.

The point is that now the portion of the double integral with $x,y\in I_j$ for fixed $j$ is only $1/4$. There are two of these, so the overall contribution is $1/2$, but that still improves by a factor of $2$ what I had before.

The contributions with $x\in I_j$, $y\notin I_j$ (or the other way around) can be kept small by taking $\delta$ sufficiently small.

Now we can iterate this procedure. It seems clear (but I haven't proved it formally) that my future antics on the $I_j$'s will not dramatically change the contributions discussed in the previous paragraph, so each step should gain me roughly a factor of $2$.

It is well known that the $H^{1/2}( R)$ norm does not control the $L^\infty$ norm. This means that there exists a sequence of test functions whose $H^{1/2}$ norm tends to zero, and whose maximum value is >2. Since norms are translation invariant, this gives the sequence you are looking for.

Concerning the failure of the continuous embedding $H^{1/2} \subset L^\infty$, the proof I know is indirect and I do not know explicit counterexamples, like the elementary ones e.g. for the failure of $ W^{1,n}(R^n)\subseteq L^\infty(R^n)$ (essentially $\log|x|$), but of course there may be some explicitly constructions.