Forgot password?
 Create new account
View 113|Reply 3

Composition of bounded linear operator

[Copy link]

3147

Threads

8495

Posts

610K

Credits

Credits
66173
QQ

Show all posts

hbghlyj Posted at 2023-4-25 08:23:48 |Read mode
Last edited by hbghlyj at 2023-4-27 10:41:00Proposition 2.4.
The composition $ST$ of two bounded linear operators $S ∈ L(Y, Z)$ and $T ∈L(X, Y )$ between normed spaces $X, Y, Z$ is again a bounded linear operator and we have
\[\|ST\|_{L(X,Z)}≤ \|S\|_{L(Y,Z)}\|T \|_{L(X,Y )}.\]

能找到$<$不取等的例子吗?最好是当 $X = Y = Z$ 并且 $\dim ⁡ ( X )$ 很小的例子。
观察它的证明,只需要 $S,T$ 不同处达到最大值。

Comment

有限维等价于研究 matrix norm, 相关定理一大把.  Posted at 2023-4-25 14:11

3147

Threads

8495

Posts

610K

Credits

Credits
66173
QQ

Show all posts

 Author| hbghlyj Posted at 2023-4-25 08:44:04
想了一下,应该可以取$X=Y=Z=ℝ^2$.
$S:(x,y)↦(2x,y),\|S\|=2$.
$T:(x,y)↦(x,2y),\|T\|=2$.
$ST:(x,y)↦(2x,2y),\|ST\|=2<2\times 2$.

3147

Threads

8495

Posts

610K

Credits

Credits
66173
QQ

Show all posts

 Author| hbghlyj Posted at 2023-4-26 18:29:23
Czhang271828  发表于 2023-4-25 07:11
有限维等价于研究 matrix norm, 相关定理一大把.

Wikipedia
Matrix norms are particularly useful if they are also sub-multiplicative:
\[\left\|AB\right\|\leq \left\|A\right\|\left\|B\right\|\]
Every norm on $K_{n×n}$ can be rescaled to be sub-multiplicative; in some books, the terminology matrix norm is reserved for sub-multiplicative norms.

手机版Mobile version|Leisure Math Forum

2025-4-21 01:12 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list