Suppose $F:\mathbb{R}^2\rightarrow\mathbb{R}$ is such that for all continuous paths $g:[0,1]\rightarrow\mathbb{R}^2$, $F(g(t))$ is continuous.
Is F continuous?
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What is your definition for $F$ to be continuous? – vadim123 Jun 19 '13 at 15:05
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I guess the usual one ? Relatively to the usual topologies on $\mathbb{R}$ and $\mathbb{R}^2$ – justt Jun 19 '13 at 15:05
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The standard definition (For every series Xn->X0, f(Xn)->f(X0) or forall eps>0 exists delta>0 s.t. ...) – ORBOT Inc. Jun 19 '13 at 15:10
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Every convergent sequence lies on a continuous path. – Martin Jun 19 '13 at 16:04
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Why, and how do you prove it? – ORBOT Inc. Jun 19 '13 at 16:11
2 Answers
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Yes: It suffices to show that $F(x_k)$ converges to $F(x)$ for any sequence $(x_k)$ in $\mathbb R^2$ that converges to $x$. This convergence follows form the consideration of $g$ whose image is the polygonal path along this sequence ($g(1-1/k) = x_k$, $g(1) = x$).
Stephen Herschkorn
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Yes.
Assume $F(0,0)=0$ and that $F$ is not continuous at $(0,0)$. Then there is an $\epsilon_0>0$ and a sequence $(z_n)_{n\geq1}$ with $\lim_{n\to\infty} z_n=(0,0)$ and $|F(z_n)|\geq \epsilon_0$ for all $n\geq1$. After passing to a subsequence we may assume $|z_n|\leq 2^{-n}$ for all $n\geq1$. Put $$g\left({1\over n}\right):=z_n\quad (n\geq 1), \quad g(0)=0,\quad g(-t)=g(t)$$ and extend $g$ to a piecewise linear function for $0<|t|\leq1$. Then $g:\ [-1,1]\to{\mathbb R}^2$ is a continuous path such that $$t\mapsto F\bigl(g(t)\bigr)$$ is not continuous at $t=0$.
Christian Blatter
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