1
$\begingroup$

let $T$ be a linear operator from a Banach space $X$ to Banach space $Y$.and $X=ker(T)\oplus M_1$ where $M_1$ is closed subspace of $X$.let $M$ be a closed subspce of $X$ then I want to prove that there exist a finite dimensional subspace $M_0$ such that $M=M \cap M_1 +M_0$

$\endgroup$
2
$\begingroup$

If $\dim (Ker(T))= +\infty$ the claim is false. Indeed putting $M = Ker(T)$ yeld $M\cap M_1 = \emptyset$ thus $M_0 = M$ must have infinite dimension.

Suppose that $\dim (Ker(T))<+\infty$.

From linear algebra $M = M\cap M_1 \oplus M_0$ for some $M_0$ (here $\oplus$ is only algebraic, it do not implies that $M\cap M_1$ is complemented as a Banach space). Consider the map $M\to X \to \frac X M_1$ where the first map is the inclusion and the second one is the projection. The kernel of this map is $M\cap M_1$ and the image is $M/M_1$. It follows from elementary algebra that $M/(M\cap M_1)\approx M/M_1$ which is finite dimensional since $M/ M_1 \lhd X/M_1 \approx Ker(T)$ and $\dim (Ker(T))<+\infty$. Thus $M_0\approx M/(M\cap M_1)$ is finite dimensional.

$\endgroup$
0
$\begingroup$

It is probably better to post the original problem.

If $X$ is separable Hilbert with $X=H_1\oplus H_2$ and $H_1, H_2$ both infinite-dimensional, and $T$ is the orthogonal projection onto $H_2$ then $M_1=H_2$.

Take $M = H_1$. Then $M\cap M_1=\{0\}$. Therefore $M_0$ would have to be $H_2$.

In your case, do you know something else about $T$?

$\endgroup$
  • $\begingroup$ In my question I am considering problem in banach space but u are considering hilbert space.@olivia $\endgroup$ – Manu Rohilla Jul 9 '17 at 4:29
  • $\begingroup$ @ManuRohilla Hilbert spaces are, in particular, Banach spaces. $\endgroup$ – Olivia Jul 9 '17 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.