For a closed subspace $M \subset X$, show the following statements:
- If $\dim M < \infty$, $M$ is complemented.
- If $\operatorname{codim} M (= \dim X/M) < \infty$, M is
complemented.
I know that to solve the first point I should choose a basis $\{x_i\}_{i=1}^n$ of $M$ and basis $\{ x_i^*\}_{i=1}^n$ of $M^*$, and extend $x_i^*$ to $\psi_i \in X^*$ by the Hahn-Banach theorem, and set $N= \cap_{i=1}^n \ker \psi_i$.
Edited
I proved the first point but the second point still not very easy to me. Could you please check my prove and help me with the other one.
Let $\{ x_1, \dots, x_n \}$ be a basis for $M$.
Define linear functionals $x^*_1, \dots, x^*_n : M \to \mathbb{R}$ by
$x^*_i(\sum_j^n c_j x_j) = c_i,$
where the $c_j$'s are coefficients in $\mathbb{R}$.
By Hahn-Banach theorem, we can extend $x^*_1, \dots, x^*_n$ to continuous linear transformations $\psi_1, \dots, \psi_n$ on $X$. Since $\psi_1, \dots,\psi_n$ are continuous, then for each $i \in \{1,\dots, n\}$, $\ker \psi_i$ is closed linear subspace. Now, put $N= \cap_{i=1}^n \ker \psi_i$, which is, clearly, a linear subspace, and $N$ is closed because it is the intersection of closed sets.
Finally, we must check that $X = M \oplus N$. For any given $v \in X$, we can write $v $ in the form $ v=x + y$ with $x \in M$ and $y \in N$ by taking $x = \sum_i \psi_i(x) x^*_i$ and $y = v – x$. it is obvious that $M \cap N = 0$. So $X = M \oplus N$.
Best Answer
Outline for 2: Consider $y_1,\ldots, y_n$ a base of $X/M$, the let $x_i$ be any preimage of $y_i$ under the projection $X\to X/M$. Let $N=\text{span}\{ x_1\ldots, x_n\}$.