# Closed Image + dim ker $D<\infty$

I have some trouble to understand the proof of following Lemma:

Lemma: $X$,$Y$ Banach spaces and let $D:X\to Y$ be a bounded linear operator. Following is equivalent: (i) dim kerD$<\infty$ and imD is closed (ii)$\exists$ Banach space $Z$ and a compact operator $K:X\to Z$ and a cst $c>0$ s.t. $\Vert x\Vert_X\leq c(\Vert Dx \Vert_y \Vert + Kx\Vert_Z)$

And in the proof by a lemma there exist bounded linear functionals $x_1^{\ast},...,x_m^{\ast}\in X^{\ast}$ s.t. $<x_i^{\ast},x_j>=\delta_{ij}$. we define the bounded operator $K:X\to Z:=kerD$ by $Kx:=\sum_{i=1}^m <x_i^{\ast},x>,x_i$, then K is compact.

Question is: why is then the operator $X \to Y\times \mathbb{R}^m:x\mapsto(Dx,Kx)$ injective?

I suppose here we are trying to prove (i) implies (ii), so we are assuming that $\dim \ker D < \infty$. Then I guess $\{x_1, \dots, x_m\}$ is supposed to be a basis for $\ker D$.
Suppose $x$ is in the kernel of the given operator, so $Dx=0$ and $Kx=0$. Since $Dx=0$ we have $x \in \ker D$ so there are unique coefficients $a_1, \dots, a_m$ with $x = \sum_{i=1}^m a_i x_i$. Then $x_i^*(x) = a_i$ so $Kx = \sum a_i x_i = x$. Since we assumed $Kx=0$ we have $x=0$. So the given operator is injective.
• "I suppose here we are trying to prove (i) implies (ii), so we are assuming that dimker$D<\infty$ Then I guess ${x_1,…,x_m}$ is supposed to be a basis for ker$D$." Yes exactly - sorry for not precifying this! Thank you Nate Elredge it helps a lot! – MorganeMaPh Dec 31 '15 at 8:40