# Hahn-Banach theorem corollaries

Notes :Corollary: Let $L\subset X$ be a Banach space, then there exists a complement of $L$ that is closed and can be defined by linear functionals linearly independent.

Proof: Consider $e_1,e_2...e_n$ a basis of $L$. Then $f_i(e_k)=\delta_{ik}$ delta Kronecker.

$x=x_1+x_2$ and $x_1=\sum_\limits{i=1}^{n}f_i(x_1)e_i$

Therefore

$x_2=x-x_1=x-\sum_\limits{i=1}^{n}f_i(x_1)e_i$

Consider $x_2\in M$ $0=x_1+x_2$ and $x_1\in L,\:\:x_2\in M$

$x_2=\sum_\limits{i=1}^{n}c_ie_i$

$c_k=f_k(x_2)=0$ and $x_1=x_2=0$

Then $L\oplus M$

Question:

1) I stop understanding at the point the author states $0=x_1+x_2$. How does the author goes from $0=x_1+x_2$ to $L\oplus M$? What is the logic behind $0=x_1+x_2$?

2) After proving the Hahn-Banach extension theorem. I am going through a lot corollaries about functionals like this one. However I have not found them in the literature within my grasp. Perhaps because these corollaries are called by another name. Could someone point me in the direction of finding these corollaries?

• Is this the proof you've found word-for-word? – Aweygan May 5 '18 at 15:02
• @Aweygan This is a notebook proof from a functional analysis class. I added the words to make it more understandable. Thanks for your comment! – Pedro Gomes May 5 '18 at 15:04
• You can find a better, more complete proof of this statement here. This particular corollary is in Rudin's Functional Analysis. Note that every Banach space is a locally convex TVS. – Michael Lee May 5 '18 at 15:16

I believe what's happening at the point where you get confused is that we assume $x_1\in L$, $x_2\in M$, and $x_1+x_2=0$. If you can then show that $x_1=x_2=0$, then it follows that $L\cap M=\{0\}$. And assuming you've already shown that $L$ and $M$ are closed and $X=L+M$, then you have $X=L\oplus M$.
• Could you please tell me $f_k(x_2)=0=c_k$ but here the author meant $f_i(x_2)=0=c_i$. I mean the functionals that acted in $L$, right? – Pedro Gomes May 5 '18 at 15:50