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Edit: After the answer of Prof. Eremenko to the previous version, I realized that a weaker assumption works for the main motivation of this post. so I revise the question.

The unit disk in the plane is denoted by $\mathbb{D}=\{(x,y) \in \mathbb{R}^2\mid x^2+y^2 <1\}$

Assume that $f:\mathbb{D} \to \mathbb{R}$ is a smooth (or real analytic function) such that for every analytic curve $\alpha \subset \mathbb{R}^2$ , $f$ is uniformly continuous on $\alpha \cap \mathbb{D}$.

Does this imply that $f\in L^2(\mathbb{D})$, or at least $f\in L^p(\mathbb{D})$for some $p >1$?

Motivations:

I encountered such type of functions in the "proof" of the proposition of the following note:

https://arxiv.org/abs/math/0408037

In reality, the proof is not correct, because the above assumption does not imply that $f$ has a (unique) continuous extension to the boundary. The mistake of the note was that it assumed such continuous extension.

More precisely we have a vector field $X=P\partial_x+Q\partial_y$ on the plane. We define a linear operator on the space of smooth functions on the plane with $L_X(f)=X.f=Pf_x+Qf_y$. We wish to solve the differential equation $$X.f=g$$.(This equation is called "Homological equation" in the following talk: "https://cms.math.ca/Events/Toulouse2004/abs/ss7.html#lt")

If $\gamma$ is a hyperbolic limit cycle of $X$ with period $T$ and surrounds a hyperbolic singularity, we can solve the above equation in the interior of $\gamma$ provided $\int_0^T g(\gamma(t))dt=0$. However we wish to pass from the interior of $\gamma$ to its exterior.

The true fact is that, in the interior of $\gamma$, the solution $f$ of $X.f=g$ satisfies the above Assumption. (In this post , for simplicity, we replace an arbitrary $\gamma$ by the unit circle) So the question in this post would be a motivation to consider a Sobolov space for action of $L_X$.

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No, it does not. Take a curve tangent to $S$ from inside, for example an arc of a circle. Then take a narrow neighborhood $V$ of this curve. Then there exists a function, $C^\infty$ in the open disk, zero outside $V$ and growing as much as you want in $V$, so it does not belong to any $L^p$.

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  • $\begingroup$ Thank you very much for your answer. Could you please give a comment on the following revised version of the assumption: for every analytic curve $\alpha$, f is uniform continuous on $\alpha \cap \mathbb{D}$. This new version work for the same motivation in my question.Moreover is there an example of real analytic f of your example? $\endgroup$ – Ali Taghavi Jun 6 '17 at 22:23
  • $\begingroup$ In my example, $f$ can be real analytic, in fact of the form $|g(z)|$ with $g$ complex analytic. If you want to modify your question, modify it in the text of the question, not in the comment. $\endgroup$ – Alexandre Eremenko Jun 6 '17 at 23:30

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