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

Question

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?

Answer 1

(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.