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The following problem is taken from 'Real Mathematical Analysis' by Pugh, $2$nd edition, page $366$, exercise $3.$

Question: Let $T:V \rightarrow W$ be a linear transformation between normed spaces. Show that $$\| T \| = \sup\{ |Tv|: |v|<1 \} = \sup\{ |Tv|: |v| \leq 1| \} = \sup\{ |Tv|: |v| =1 \} = \inf\{ M:v \in V \Rightarrow |Tv| \leq M |v| \}.$$

Definition of $\| T \| = \sup \{ \frac{|Tv|}{|v|}: v \neq 0 \}.$

I managed to show that $\| T \| \leq \sup\{ |Tv|: |v \leq 1| \} \leq \sup\{ |Tv|: |v| =1 \} \leq \inf\{ M:v \in V \Rightarrow |Tv| \leq M |v| \} \leq \| T \|.$

However, I have no idea on how to show $\| T \| \leq \sup\{ |Tv|: |v|<1 \}.$

Any hint would be appreciated.

Idonknow
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  • This question has been asked here: https://math.stackexchange.com/questions/171259/equivalent-definitions-of-the-operator-norm – Skyhit2 Jun 09 '17 at 23:24
  • @Skyhit2: No, my question involves $|v|<1,$ a strict inequality. the link you posted does not contain such inequality. – Idonknow Jun 09 '17 at 23:25
  • $T$ is continuous, so if you have $v_n \rightarrow v$ with $v$ on the unit sphere then either $|Tv_n| \geq |Tv|$ for some $n$ and the statement is proven by definition of $\sup$, or otherwise $\sup \lbrace |Tv_n| \rbrace = |Tv|$ and by continuity of $T$ (and absolute value) in fact $|Tv_n| \rightarrow |Tv|$, but then $|Tv| \leq \sup \lbrace |Tv_n| \rbrace \leq \sup \lbrace |Tv| : v \in \text{unit ball} \rbrace$. – John Samples Jun 09 '17 at 23:29
  • @Idonknow: Since we're taking the supremum the strict and not strict inequalities are equivalent. e.g. $\text{sup}{x \in \mathbb{R} \colon x^2 < 2} = \text{sup}{x \in \mathbb{R} \colon x^2 \leq 2}$. – Skyhit2 Jun 09 '17 at 23:30
  • this is only true when $T$ is continuous. This happen when $V$ is finite-dimensional or when $T$ is a bounded map in the case of $V$ infinite dimensional. – Masacroso Jun 09 '17 at 23:40
  • Oh, I didn't notice this. Well, by continuity on a one-dimensional subspace you can use the same argument as I mentioned; for $v$ on the unit sphere, just take $v_n = (1 - \frac{1}{n})v$. – John Samples Jun 10 '17 at 00:03

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