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Assume that $T$ is a linear transformation from $V$ to $V'$ (which are two vector spaces having the inner products $\|\cdot\|$ and $\|\cdot\|'$ respectively)
Then, The operational norm of $T$ is called $\|T\|_{op}$ and defined like this:

$\|T\|_{op} := \sup\{\|T(x)\|':\|x\| \le 1\}$


Assume that $V$ is a vector space and two norms $\|\cdot\|$ and $\|\cdot\|'$ are defined on $V$. We say $\|\cdot\|$ and $\|\cdot\|'$ are equivalent if there exist $M,m \gt 0$ such that:
$\forall x \in V \space\space m\|x\|\le \|x\|' \le M\|x\|$


Assume that $V$ is a vector space which is not necessarily of finite dimension and $\|\cdot\|$ and $\|\cdot\|'$ are two norms on it. If the identity linear transformation has finite operational norm in these two situations written below, Prove that $\|\cdot\|$ and $\|\cdot\|'$ are equivalent norms.

$1_V:(V,\|\cdot\|') \to (V,\|\cdot\|)$ and $1_V:(V,\|\cdot\|)\to (V,\|\cdot\|')$

Note: The question seems too complex! I can't even understand it! I'm completely blind! I'm given few things and asked a lot!

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  • $\begingroup$ Did you try to start writing down the precondition that the two given maps have finite operator norm explicitely? $\endgroup$ – Fabian Apr 13 '17 at 16:56
  • $\begingroup$ @Fabian excuse me sir, what do u mean by "writing down"? i apologize if i'm stupid but isn't it written above? $\endgroup$ – Arman Malekzadeh Apr 13 '17 at 17:08
  • $\begingroup$ It is written for a general operator $T$. You should write it down for the two operators $1_V$. In particular, what is the definition of $\sup$. Once you have written this down, it should be obvious how you can find $m$ and $M$ as required. $\endgroup$ – Fabian Apr 13 '17 at 17:10
  • $\begingroup$ @Fabian you mean, finiteness means having maximum value here? $\endgroup$ – Arman Malekzadeh Apr 13 '17 at 17:23
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For any $x\in V$, \begin{equation*} \|x\| = \|Ix\| \leq \|I\|_{op}\|x\|', \end{equation*} Where I have used $\|I\|_{op}$ to denote the operator norm of $I:(V, \|\cdot\|') \to (V, \|\cdot\|)$. The reverse inequality is similar but you have to use the operator norm of $I:(V, \|\cdot\|)\to (V, \|\cdot\|')$.

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  • $\begingroup$ Why is $||Ix|| \le ||I||_{op} ||x||'$ true? why not $||Ix|| \le ||I||_{op} ||x||$? $\endgroup$ – Arman Malekzadeh Apr 14 '17 at 12:26
  • $\begingroup$ The inequality I've written is true by the assumption that when viewed with domain $(V, \|\cdot\|')$ and codomain $(V, \|\cdot \|)$, $I$ is bounded. Of course the inequality you've written is also true, but it isn't helpful. $\endgroup$ – BindersFull Apr 14 '17 at 13:43

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