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Show that

$id: (\mathbb{R}^n, ||.||_2)\rightarrow (\mathbb{R}^n, ||.||_X): x \mapsto x$

is a continuous function. $||.||_X$ is any Norm of $\mathbb{R}^n$.

I don't know how to deal with this kind of Problems. How to show that a function with Norms is continuous?

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    Hint: $\mathbb{R}^n$ is a finite-dimensinal vector space, then it admits a finite basis. – 2017-01-15

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If you do not know that all norms on $\mathbb R^n$ are equivalent, you can argue as follows:

write $x=x_1\vec e_1+\cdots +x_n\vec e_n$ where $\left \{ \vec e_i \right \}_i$ is the standard basis on $\mathbb R^n$ and $x_i\in \mathbb R$. Then, using the triangle inequality followed by Cauchy-Schwarz, we have

$\left \| x \right \|_{X}\le |x_1|\cdot \left \| \vec e_1 \right \|_{X}+\cdots +|x_n|\cdot \left \| \vec e_n \right \|_{X}\le \left \| x \right \|_{2}\sqrt{\left \| \vec e_1 \right \|_{X}^{2}+\cdots +\left \| \vec e_n \right \|_{X}^{2}}$,

so with $C=\sqrt{\left \| \vec e_1 \right \|_{X}^{2}+\cdots +\left \| \vec e_n \right \|_{X}^{2}}$, we find that $\left \| x \right \|_{X}\le C\left \| x \right \|_{2}$, from which it is almost immediate that $id$ is continuous.

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    Thank you! Can you please explain why it is immediate that id is continuous if the inequality holds? – 2017-01-15
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    Because $||id(x)-id(x_0)||_X=||x-x_0||_X\le C||x-x_0||_2$.Now let $x\to x_0$ in the 2-norm. i.e. for a given $\epsilon$ challenge taking $\delta =\epsilon /C$ will do the trick. – 2017-01-15
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If your definition of continuity is topological then the argument is as follows.

Since all norms are equivalent in $\mathbb{R}^n$, the topologies $\tau_1,\tau_2$ induced respectively by $||\cdot||_X$ and $||\cdot||_2$ are equal. In particular $\tau_1\subset\tau_2$, as is needed to show continuity of $id$.