2
$\begingroup$

I've been struggling with this exercise and i couldnt find any valid argument :

Let $X$ a Banach Space and $Y$ a closed subspace of $X$. If $Y$ has finite dimension then Y is a complemented subspace of $X$.

Am giving all the necessary definitions :

According to my book :

A subspace $Y$ of a Banach space $X$ is called complemented subspace if there exist a linear bounded operator $p:X \mapsto Y$ such that $p(x)=x$ for all $x \in Y$.

I have shown in a previous exercise that this is equivalent to :

There exist a closed subspace $Z$ of $X$ such that $ Y \cap Z = \{0\} $ and $X = Y \oplus Z$, meaning that for all $x \in X$ there are unique $ y \in Y $ and $z \in Z$ such that $x = y +z .$

Thanks for your time !

$\endgroup$
2
$\begingroup$

Let $\{y_1,\cdots, y_n\}$ be a basis of $Y$. Now define $f_i:Y\to \mathbb{R}$ as $$f_i(r_1y_1+\cdots + r_ny_n) = r_i.$$ It is bounded linear operator over $Y$. Hence by Hahn-Banach theorem we can extend $f_i$ to the whole space. Now take $p(x) = f_1(x)y_1+\cdots f_n(x)y_n$.

$\endgroup$
  • $\begingroup$ Nice , so it had a trick ! :P $\endgroup$ – dem0nakos Apr 8 '18 at 10:02

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.