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.



Approximate constant function
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. 


