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#2: Post edited
by
whybecause
·
2022-01-31T02:39:42Z (almost 3 years ago)
Added duh
- Let $X$ be a normed linear space and suppose that, for each $f\in X$ there exists a bounded linear functional $ T\in X^* $ such that $T(f)=||f||$ and $||T||_*=1$. Prove that if $\\{f_n\\}\to f$ in $X$ then we have $||f||\le\liminf||f_n||$.
- ---
- I think the only $T$ that makes sense to work with is that for which $T(f)=||f||$ and $||T||_ * = 1$. Because $||T||_ * =1$ we therefore have $|T(x)| \le ||x||$ for all $x\in X$. Therefore by the continuity property of bounded linear functionals
- $$ ||f|| = |T(f)| = \lim |T(f_n)| \le \lim ||f_n|| $$
It seems like I must have made a mistake because I seem to have proved a result stronger than the problem asked for. But I don't see any invalid step.
- Let $X$ be a normed linear space and suppose that, for each $f\in X$ there exists a bounded linear functional $ T\in X^* $ such that $T(f)=||f||$ and $||T||_*=1$. Prove that if $\\{f_n\\}\to f$ in $X$ then we have $||f||\le\liminf||f_n||$.
- ---
- I think the only $T$ that makes sense to work with is that for which $T(f)=||f||$ and $||T||_ * = 1$. Because $||T||_ * =1$ we therefore have $|T(x)| \le ||x||$ for all $x\in X$. Therefore by the continuity property of bounded linear functionals
- $$ ||f|| = |T(f)| = \lim |T(f_n)| \le \lim ||f_n|| $$
- It seems like I must have made a mistake because I seem to have proved a result stronger than the problem asked for. But I don't see any invalid step.
- ---
- Sorry, no I didn't, and duh. I proved something weaker than what the problem asked for. Would still appreciate insight into how to prove this theorem.
#1: Initial revision
by
whybecause
·
2022-01-31T02:27:52Z (almost 3 years ago)
$\{f_n\}\to f$ in $X$ implies $||f||\le\liminf||f_n||$
Let $X$ be a normed linear space and suppose that, for each $f\in X$ there exists a bounded linear functional $ T\in X^* $ such that $T(f)=||f||$ and $||T||_*=1$. Prove that if $\\{f_n\\}\to f$ in $X$ then we have $||f||\le\liminf||f_n||$.
---
I think the only $T$ that makes sense to work with is that for which $T(f)=||f||$ and $||T||_ * = 1$. Because $||T||_ * =1$ we therefore have $|T(x)| \le ||x||$ for all $x\in X$. Therefore by the continuity property of bounded linear functionals
$$ ||f|| = |T(f)| = \lim |T(f_n)| \le \lim ||f_n|| $$
It seems like I must have made a mistake because I seem to have proved a result stronger than the problem asked for. But I don't see any invalid step.