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Assume we have a function $f:\Omega_1\times\Omega_2\rightarrow \mathbb{C}$ that is bounded and differentiable w.r.t to the second variable. Further let $\Omega_2$ be bounded. The derivative $\partial_{\omega_2}f$ is known to be continuous in both variables. Is it now true, that $\partial_{\omega_2}f$ is bounded?

I think not but i can't find an counter example.

Edit: Of course $\Omega_1$ is unbounded in general.

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  • $\begingroup$ Take $\Omega_1 = \Omega_2 = (-1,1)$ and $f(x,y) = \sqrt{1-y^2}$. $\endgroup$ – user38355 Sep 16 '13 at 10:13
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If you don't require continuity of $\partial_{\omega_2}$ up to the boundary, there are counterexamples even when $\Omega_1$ is bounded, see the comments.

Here's another example that works on $\mathbb R \times \mathbb R$ (as well as if you replace the second $\mathbb R$ with something bounded): $$f(\omega_1, \omega_2) = \sin (\omega_1 \omega_2)$$

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