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Suppose $\;F:\Bbb R^2\to\Bbb R\;$ is such that for any continuous path $\;\gamma:[0,1]\to\Bbb R^2\;$ , the composition $\;F\circ \gamma:[0,1]\to\Bbb R\;$ is continuous . Is then $\;F\;$ continuous? The remarked continuous above is part of the question, not of me.

Now, I'm very stuck in this though I'm pretty sure the claim is false, but I haven't got any counter example. "Usual" examples of functions discontinuous at the origin don't seem to work, and I think maybe some weird way to pass to the limit (at origin or else where) could help here.

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The claim is true.

Assume $F$ is not continuous, so without loss of generality, it is discontinuous at $0$, and $f(0)=0$. Thus there is some $\epsilon>0$ such that for each $\delta>0$ there is some $x\in B(0,\delta)\subset\mathbb{R}^2$ with $|F(x)|>\epsilon.$ For any natural $n$ let $x_n\in B(0,1/2^n)$ such that $|F(x)|>\epsilon$. Let $c$ be the infinite polygon that connects every $x_n$ to $x_{n+1}$ by the straight segment between them. By construction, $c$ is of finite length, and obviously it gets closer to the origin as $n\to\infty.$ Now take $\gamma$ to be the arc length parametrization of $c$, with the origin as its endpoint. This is indeed a continuous path, and the composition $F\circ\gamma$ is not continuous.

Amitai Yuval
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  • Why "by construction" the "c" is finite? There are infinite paths bounded in the plane, aren't there? Thank you for the posting. –  Oct 18 '14 at 18:33
  • By construction, the sequence $|x_n-x_{n+1}|$ is bounded by a geometric sequence with quotient=$1/2$, thus the total length is bounded. – Amitai Yuval Oct 18 '14 at 18:37