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So I don't have a whole lot of experience in general proving continuity for multivariable functions, and I want to make sure I'm going about things correctly.

Prove that the function $B(z,w):=\int_0^1 t^{z-1}(1-t)^{w-1}dt$, for $z,w\in \mathbb{C}$, is continuous.

So I let $\varepsilon > 0$ and basically consider an expression of the form:

$$|\int_0^1 t^{z-1}(1-t)^{w-1}dt - \int_0^1 t^{u-1}(1-t)^{v-1}dt| < \varepsilon$$

Which simplifies into the form:

$$|\int_0^1 t^{z-1}(1-t)^{w-1}[1-t^{u-z}(1-t)^{v-w}]dt| < \varepsilon$$

And now to make this expression true I consider $|u-z|<\delta_1$ and $|v-w|<\delta_2$ and choose $\delta_1$ and $\delta_2$ as small as I need to (based on $\varepsilon$), which will allow me to make this integral as small as a wish, since $t^{u-z}(1-t)^{v-w}\rightarrow 1$ as $\delta_1,\delta_2 \rightarrow 0$.

Is this an acceptable way to go about continuity in several complex variables, should only a single delta equal to $Min[\delta_1,\delta_2]$ be used, are there any other subtleties I should be made aware of in the multivariable case?

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1 Answer 1

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when you are talking about continuity you should know your topology. your domain is $\mathbb{C} \mathrm{x} \mathbb{C}\, $ your topology is induced by your norm in your case your norm could be $| \! | (u,w | \! | = |u|+|w| $ or $\| \! | (u,w) | \! | = \sqrt{|u|^2+|w|^2} $ so normally by definition for $ \epsilon > 0$ you should find a $\delta >0 $ such that for $| \! | (u,w)- (u_0,w_0) | \! | |< \delta \Rightarrow |\!|B(u,w)-B(u_0,w_0)|\!|< \epsilon $. So I think now, using these norms what you did is equivalent.

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