Let $f:[0,1]^2 \rightarrow \mathbb C$ be an $H^1$ function with the property that $f(x,x)=0$ and $\Vert f \Vert_{L^2[0,1]}=1.$
Does there exist a constant $c>0$ such that any such function satisfies
$$ \Vert f-1 \Vert_{H^1}>c?$$
I was thinking that the Fourier series could help to prove or disprove something like this, but I did not get far so far.
It would be clearly possible in $L^2$ norm let's say, but I find it tricky in Sobolev norms.
For all $x$ and $y$ in $[0,1]^2$
$$f(x,y)=
\left\{
\begin{aligned}
\int_x^y f_y(x,z)\,dz&\text{ if }x\le y, \\
-\int_y^x f_y(x,z)\,dz&\text{ if }x\ge y,
\end{aligned}
\right.
$$
where $f_y(x,z):=\frac{\partial f(x,y)}{\partial y}|_{y=z}$, so that
$$|f(x,y)|\le\int_0^1|f_y(x,z)|\,dz\le\sqrt{\int_0^1|f_y(x,z)|^2\,dz}.
$$
Hence,
$$
\begin{split}
1 & = \iint_{[0,1]^2}dx\,dy\,|f(x,y)|^2 \\
& \le \iiint_{[0,1]^3}dx\,dy\,dz\,|f_y(x,z)|^2\\
& = \iint_{[0,1]^2}dx\,dz\,|f_y(x,z)|^2
\le\|f-1\|_{H^1}^2.
\end{split}
$$
So, any $c\in(0,1)$ will do.
