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Let $X$ be a Banach space, $Y$ a normed space, $A: X \to Y$ linear operator. Show that $A$ is bounded if and only if norm $||x||_A=||x||+||Ax||$ is complete.

Proof $\implies)$ $$ ||x_m-x_n||_A=||x_m-x_n||+||Ax_m-Ax_n||\le||x_m-x_n||(||A||+1)\le \epsilon(||A||+1)$$

But does it get me anywhere nearer the fact that this norm is complete?

Pedro
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maq
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  • We have also $|x_m-x_n|\le|x_m-x_n|_A$. So $|\cdot|_A$ Cauchy $\Rightarrow$ $|\cdot|$ Cauchy $\Rightarrow$ $|\cdot|$ converges $\Rightarrow$ ... – A.Γ. May 24 '17 at 21:59
  • Every Cauchy sequence w.r.t. the new norm converges in the original norm (because such a sequence is also Cauchy w.r.t. the original norm). From the boundedness we conclude that the sequence converges in the new norm. See my post for details. – Pedro May 24 '17 at 22:07

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$(\Rightarrow)$ Let $(x_n)$ be Cauchy w.r.t. $\|\cdot\|_A$. Then $(x_n)$ is Cauchy w.r.t. $\|\cdot\|_X$ because $\|\cdot\|_X\leq \|\cdot\|_A$. Thus, there is $x\in X$ such that $\|x_n-x\|_X\to 0$ (because $X$ is Banach). As $A$ is bounded, it follows that $\|Ax_n-Ax\|_Y\to 0$. This implies that $x_n\to x$ w.r.t. $\|\cdot\|_A$. So, $\|\cdot\|_A$ is complete.

Remark. The fact that $x_n\to x$ w.r.t. $\|\cdot\|_A$ follows from your estimate with $x_m$ replaced by $x$. This is the main point: the boundedness of $A$ should be used to prove that convergence in $\|\cdot\|_X$ implies convergence in $\|\cdot\|_A$. So, we have to deal with the diference $x_n-x$ instead of $x_m-x_n$ (as in your trying). The fact that Cauchy in $\|\cdot\|_A$ implies Cauchy $\|\cdot \|_X$ follows from the definition of the norm and thus it has nothing to do with the boundeness.

Remark 2: The converse follows from the Open Mapping Theorem (see here): $$\|Ax\|_Y\leq \|x\|_A\leq c\|x\|_X.$$

Pedro
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