# How can I prove that the Hilbert transform on the 1-torus doesn't map $L^1(\mathbb{T})$ into itself?

Let $$\mathbb{T}$$ be the 1-torus. Then, it is well defined the Hilbert transform: $$\mathcal{H}:L^1(\mathbb{T})\to L^0(\mathbb{T}), \vartheta\mapsto\int_{-\pi}^\pi f(\vartheta-t)\cot\left(\frac{t}{2}\right) \frac{\operatorname{d}t}{2\pi}.$$ I know that $$\mathcal{H}$$ is weak-(1,1), i.e. that there exists a constant $$C>0$$ such that: $$\forall f\in L^1(\mathbb{T}), \forall \lambda>0, \left|\left\{\vartheta\in\mathbb{T} : |\mathcal{H}(f)(\vartheta)|>\lambda \right\}\right|\le C\frac{\|f\|_1}{\lambda}.$$ In my lecture notes it is stated without proof that $$\mathcal H$$ doesn't map $$L^1(\mathbb{T})$$ into itself.

I'm thinking about how to prove this claim.

I can prove that doesn't exist an operator $$\mathcal{G}: L^1(\mathbb{T})\to L^1(\mathbb{T})$$ such that $$\forall f\in L^1(\mathbb{T}), \forall n\in\mathbb{Z}, \mathcal{F}(\mathcal{G}(f))(n) = -i \operatorname{sgn}(n)\mathcal{F}(f)(n),$$ where $$\mathcal{F}$$ is the Fourier transform.

Also, I know that $$\forall p\in(1,+\infty), \forall f\in L^p(\mathbb{T}), \left(\mathcal{H(f)}\in L^p(\mathbb{T})\right)\land \left(\mathcal{F}(\mathcal{H}(f))(n) = -i \operatorname{sgn}(n)\mathcal{F}(f)(n)\right).$$

Now, if I only could prove that "$$\mathcal{H}$$ maps $$L^1(\mathbb{T})$$ into itself" would imply "$$\mathcal{H}: L^1(\mathbb{T})\to L^1(\mathbb{T})$$ with continuity", then the claim that $$\mathcal H$$ doesn't map $$L^1(\mathbb{T})$$ in itself would follows from a density and continuity argument.

However I can't see how to prove that the only fact that $$\mathcal{H}$$ maps $$L^1(\mathbb{T})$$ into itselt would imply its continuity (I thought to use something like the closed graph theorem, but I can't see how it could apply here).

Any help?

(Using a plain "$$H$$" for the Hilbert transform; easier to type and also more standard):

Yes, $$\newcommand{\sgn}{\operatorname{sgn}}$$

Prop. If $$H(L^1(\Bbb T))\subset L^1(\Bbb T)$$ then $$H:L^1(\Bbb T)\to L^1(\Bbb T)$$ is bounded.

And yes, this is immediate from the Closed Graph Theorem.

Say $$m(k)=-i\sgn(k)$$, so $$H$$ is characterized by $$\widehat{Hf}(k)=m(k)\hat f(k).$$

Suppose $$f_n\to f$$ in $$L^1$$ and also $$Hf_n\to g$$ in $$L^1$$; we need to show that $$g=Hf$$. But $$\widehat g(k)=\lim_n\widehat{Hf_n}(k)=\lim_n m(k)\widehat {f_n}(k)=m(k)\widehat f(k).$$

Note If we're talking about $$L^1(\Bbb R)$$ instead of $$L^1(\Bbb T)$$ then the Fourier transform makes it trivial to give a specific counterexample:

Triviality If $$f\in L^1(\Bbb R)$$ and $$\int f\ne0$$ then $$Hf\notin L^1(\Bbb R)$$.

Proof: Since $$\widehat f(0)\ne0$$ the function $$m\widehat f$$ is not continuous at the origin, so there cannot exist $$g\in L^1(\Bbb R)$$ with $$m\widehat f=\widehat g$$.

Edit: I tend to think of $$H$$ as defined by the Fourier transform. If we take it to be defined as the pointwise principal-value integral that doesn't quite work for $$L^1$$, because it's not at all clear that $$Hf$$ is something that even has a Fourier transform.

I thought I had a simple answer to that but it was wrong. Will give the simple answer if I figure it out.

• Thanks for the answer. However I can't see how I can prove that if $H$ maps $L^1$ into itself, then for $f\in L^1(\mathbb{T})$, the integral definition via principal value (that is the definition for Hilbert transform that I use here) of $Hf$ has a Fourier transform that satisfies $\widehat{Hf}(k)=-i\operatorname{sgn}(k)\hat{f}(k)$. I know that this relation is valid for $f\in L^p$ if $p>1$, but here $p=1$. How can I show that this relation is still valid for $p=1$? – Bob Mar 1 at 12:42