Prove identity map $i_X:(X, ||.||_1) \to (X, ||.||_2)$ is continuous, then $\exists L>0$ s.t. $\forall x \in X, ||x||_1 \le L||x||_2$

Asked Nov 01 '22 at 17:39

Active Nov 01 '22 at 17:39

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The previous version is if there exists a positive real number L s.t. $\forall x \in X, ||x||_1 \le L||x||_2$, then $i_X:(X, ||.||_1) \to (X, ||.||_2)$ is continuous.

My proof is following \begin{align*} \forall \alpha \in X, \forall \epsilon > 0, \exists \delta>0, \delta < \epsilon/k \\\\ ||x-\alpha||_1 <\delta &\Rightarrow ||i(x)-i(\alpha)||_2 \\\\ &=||x-\alpha||_2<L||x-\alpha||_1 < L\delta<\epsilon \end{align*}

So the function is continuous. But the converse version is very confusing, I'm not sure if I should use contradiction or direct prove, can someone give a hint?

asked Nov 01 '22 at 17:39

BlackTea

1/3 What is $X$ (and $|~|_p$ on it)? Is it any real vector space and any two norms? this should be specified in your post. 2/3 The indices of your two norms in the inequality must be exchanged for your title, claim, and proof to be correct. 3/3 Delete this $\delta<\epsilon/k$ in the middle of your proof. – Anne Bauval Nov 01 '22 at 17:48
Does this answer your question? Prove that if a linear operator is continuous, then it is bounded. – Anne Bauval Nov 01 '22 at 18:00
Yes, that is very helpful. – BlackTea Nov 01 '22 at 22:48

Prove identity map $i_X:(X, ||.||_1) \to (X, ||.||_2)$ is continuous, then $\exists L>0$ s.t. $\forall x \in X, ||x||_1 \le L||x||_2$

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