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$f$ is a function from $\mathbb{R}^2\to \mathbb{R}$. I feel like this can't be true but I need to come up with some counterexample, but I'm not sure how. Can you please help?

able20
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  • What do you mean by being $C^\infty$ in $x$ and $y$ respectively, in $x$, and in $y$? – R.V.N. Jan 30 '21 at 22:09
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    Do you mean $(x\mapsto f(x, y)), (y\mapsto f(x,y))\in C^\infty(\mathbb{R},\mathbb{R)}$ and want to know if $f\in C^\infty(\mathbb{R}^2,\mathbb{R})$ – GhostAmarth Jan 30 '21 at 22:11
  • Yes, I have a function specifically sending $(x,z)$ to $(-\sqrt{1-x^2-z^2}, z)$ and want to show that it is smooth. Second coordinate is trivial, but the first coordinate in the codomain is $C^{\infty}$ in $x$ and $C^{\infty}$ in $z$ but to see it is $C^{\infty}$, I need to know it is $C^{\infty}$ simultaneously... – able20 Jan 30 '21 at 22:11
  • I should have phrased that way, thanks for the clarification – able20 Jan 30 '21 at 22:14
  • look at this https://math.stackexchange.com/questions/4006005/if-each-slice-of-f-x-times-y-rightarrow-z-is-continuous-then-f-is-continu/4006081#4006081. it answers a different question but ita can be applied here as well. – alphaomega Jan 30 '21 at 22:16
  • i think the example of alphaomega should work. i was mistaken... – GhostAmarth Jan 30 '21 at 22:42
  • Thank you very much. It seems I have to get my hands dirty because I cannot use general theory. – able20 Jan 30 '21 at 22:46

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