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?


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 $x=0$.

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  • $\begingroup$ 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

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