Linear operators mapping finite dimensional vector spaces are bounded,

Theorem: Every linear operator $T:V\to W$ , where $V$ is finite dimensional, is bounded.

Proof $\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{c(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{c}$

where $\|T(e_i)\|=\max\{\|T(e_1)\|,\|T(e_2)\|,\dots\}$ .

What we learn from here is

$\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq c(|a_1|+|a_2|+\dots+|a_n|)$

where

$\|e_i\|=\max\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}$ .

Similarly,

$\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq \|e_k\|(|a_1|+|a_2|+\dots+|a_n|)$

where

$\|e_k\|=\min\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}$

Another proof of the assertion is

$\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|e_k\|(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{\|e_k\|}$

which is a constant.

Note: why does this not work in infinite dimensional spaces? Because the max and min of $\|e_r\|$ and $\|Te_r\|$ might not exist.

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Cancel

Connecting to %s

Share this:

Like this:

Related

Published by ayushkhaitan3437

Leave a Reply Cancel reply