2
$\begingroup$

Given a Banach space $(X,\|,\|)$, and a finite dimensional subspace $F \subset X$, is it always possible to choose a closed linear complement to $F$. Explicitly, I mean to say, will there always exist a closed linear subspace $K \subset X$, such that $$ X \simeq F \oplus W? $$ If yes, then what is the easiest way to see that this is the case?

$\endgroup$
3
$\begingroup$

Let $e_1,..,e_n$ be a basis of $F$, consider $f_i$ on $vect(e_1,...,e_n)$ by $f_i(e_i)=1, f(e_j)=0, j\neq i$, you can extend it to $X$ by Hahn Banach. Write $W=\cap_{i=1,..,n}Ker g_i$.

$W$ is closed, if $x\in X, x=g_1(x)e_1+..+g_n(x)e_n+(x-g_1(x)e_1+..+g_n(x)e_n)$,

write $y=x-(g_1(x)e_1+..+g_n(x)e_n)$, $g_i(y)=0$, and $x\in Vect(e_1,...,e_n)\cap W$ implies that $x=x_1e_1+...+x_ne_n$, and $g(x)=x_i=0$, we deduce that $x=0$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What is $g_i$? I guess you mean $f_i$? $\endgroup$ – Dave Shulman May 13 '19 at 17:01
  • $\begingroup$ $f_i$ is defined on $Vect(e_1,...,e_n)$ one needs Hahn Banach to extend it to $X$ in $g_i$. $\endgroup$ – Tsemo Aristide May 13 '19 at 17:05

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.