# Approximate constant function

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.

• Should $\|f\|_{L^2[0, 1]}$ be $\|f\|_{L^2[0, 1]^2}$? Jan 15 '20 at 14:52

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.