Prove that the product topology in $\Bbb C^n$ is equal to the usual one

Question

So it is well know that the function $\tau_n:\Bbb C^n\times\Bbb C^n\rightarrow\Bbb C$ defined by the condition

1. $\tau_n(x,y):=\sum_{i=1}^n x_i\overline y_i$

for any $x,y\in\Bbb C^n$ is an inner product. So I ask to prove that the product topology on $\Bbb C^n$ induced by the inner product $\tau_1$ is equal to the topology $\tau _n$ as above defined. I point out that I need of this result to show that the linear functions between two topological vector spaces are continuous and so to show that all topologies in a finite dimensional topological vector space are equivalent and thus I courteously request to not give what just said as answer. So could someone help me, please?

Answer 1

The product topology is generated by the norm

$$N_\infty(x)=\max(\vert x_1\vert, \dots \vert x_n\vert)$$ where $\vert x \vert = \sqrt{\tau_1(x,x)}$. Denoting

$$N_2(x) = \sqrt{\tau_n(x,x)}$$ we have

$$1/\sqrt{n}N_\infty(x) \le N_2(x) \le \sqrt{n} N_\infty(x)$$ which enables to conclude to the desired result.