If $f\in H^1(0,1)$ is bounded below, then $1/f\in H^1(0,1)$

Question

Let $f\in H^1(0,1)$ be such that $|f(x)|\ge a > 0$ for a.e. $x\in (0,1)$. I have problems showing that then $1/f\in H^1(0,1)$ by using the definition.

Of course, we know that $f$ is absolutely continuous with $L^2$-derivative and it is easy to see that also $1/f$ is absolutely continuous with derivative $-f'/f^2$ which is in $L^2(0,1)$. So, $1/f\in H^1(0,1)$. However, if I try to show this using the definition of $H^1$ with $C_0^\infty$ test functions, I awkwardly fail. Does anyone know how to do this?

BTW: I have the same problem with the square root of $f$.

Answer 1

We give a proof for $H^1(\mathbb R)$ that works just as well for $H^1(\Bbb R^d)$; probably with a little finagling you can make it work for $H^1(a,b)$.

Define $$\tau_hf(x)=f(x+h).$$

Lemma. $f\in H^1(\Bbb R)$ if and only if $f\in L^2(\Bbb R)$ and $||\tau_h f-f||_2\le c|h|$.

Comment Of course if $f\in L^2(a,b)$ then $\tau_h f$ is not even defined on $(a,b)$; to prove the lemma in that context you first need to concoct a suitable definition...

Proof: This is an easy exercise using the Fourier transform. qed.

Of course that's going to be relatively useless on $(a,b)$. It seems clear you could also argue more or less as follows - if something's not clear below say it's an outline or a sketch or a hint:

Proof, More Or Less: Suppose first $||f_h||_2$ is bounded, where $f_h=(\tau_hf-f)/h$. There is a subsequence $f_{h_n}$ that converges to $g$ weakly in $L^2$. If $\phi$ is a test function then $(\tau_{-h}\phi-\phi)/h\to-\phi'$ in $L^2$; it follows that $g$ is the weak derivative of $f$.

Otoh suppose that $f\in H^1(\Bbb R)$. Then it's easy to see that $f_h\to f'$ weakly in $L^2$. Hence uniform boundedness shows that $||f_h||_2$ is bounded for $0<h<\delta$. It is bounded for $\delta\le h\le 1$ because it depends continuously on $h$, and if $h>1$ then $||f_h||_2\le 2||f||_2$.

Proposition. If $f\in H^1(\Bbb R)$ and $F$ is Lipschitz with $F(0)=0$ then $F\circ f\in H^1(\Bbb R)$.

Proof: First, $F\circ f\in L^2$, since $|F(t)|\le c|t|$. And also $$|\tau_h (F\circ f)-F\circ f|\le c|\tau_h f-f|,$$hence $$||\tau_h (F\circ f)-F\circ f||_2\le c||\tau_h f-f||_2.$$

Hmm, I bet the following version of the Lemma works on $(a,b)$:

Conjecture $f\in H^1(a,b)$ if and only if $f\in L^2(a,b)$ and $\int_{I_h}|\tau_h f-f|^2\le ch^2$, where $I_h$ is the subset of $(a,b)$ on which $\tau_h f$ is defined.

If true, the proof of the Proposition works without significant change.