Let $T>0$ and $d\in\mathbb N$. How can we prove that $$C^0([0,T],\mathbb R^d)\times[0,T]\to\mathbb R^d\;,\;\;\;(\omega,t)\mapsto\omega(t)$$ is (jointly) continuous? That seems to be an easy task. So easy, that I don't see how can I prove it.
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Which topology are you considering on $C^0([0,T],\mathbb{R}^d)$? [The question is more which description of the topology is used, it's to be expected that the topology of uniform convergence = the compact-open topology is used. But which characterisation of that topology?] – Daniel Fischer Sep 27 '16 at 10:31
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@DanielFischer I assume that $C^0([0,T],\mathbb R^d)$ is equipped with the topology generated by the supremum norm $$\left|\omega\right|\infty:=\sup{[0,T]}\left|\omega\right|;;;\text{for }\omega\in C^0([0,T],\mathbb R^d);.$$ – 0xbadf00d Sep 27 '16 at 15:30
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Okay, so for every $\varepsilon > 0$ you want to find a $\delta > 0$ such that $\lVert \omega - \omega_0\rVert_{\infty} < \delta$ and $\lvert t - t_0\rvert < \delta$ together imply $\lvert \omega(t) - \omega_0(t_0)\rvert < \varepsilon$. Write $\omega(t) - \omega_0(t_0) = \bigl(\omega(t) - \omega_0(t)\bigr) + \bigl(\omega_0(t) - \omega_0(t_0)\bigr)$. – Daniel Fischer Sep 27 '16 at 15:51