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I'm trying to follow the proof of Lemma 2.1 in these notes:
Applications of Partial Differential Equations To Problems in Geometry

In particular, I'm struggling to follow the proof of the implication (a) implies (b). The set up is this:

Let $ X $, $ Y $, and $ Z $ be reflexive Banach spaces with $ X \to Y $ a compact injection and $ L : X \to Z $ a continuous linear map. Then, if the image $ L(X) $ is closed and $ \ker L $ is finite dimensional, there exists constants $ c_1 , c_2 $ such that for all $ x \in X $

$ \lVert x \rVert_X \leq c_1 \lVert Lx \rVert_Z+ c_2 \lVert x \rVert_Y $.

The proof given is: Write $ X = X_1 \oplus \ker L $ so the restriction of $ L $ to $ X_1 $ is injective. The closed graph theorem then gives the result.

I don't understand how the closed graph theorem gives the result.

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I will provide you a step by step answer, in a more talking way and maybe with some delayed explanations to allow you some thinking time. Then I will summarize the steps in a simple chain of inequalities.

1) Step by step: First look at the inequality $\|x\|_X \leq c_1 \|Lx\|_Z$. What does it mean? Well, it says that the inverse of $L$ restricted to $L(X)$, i.e.

$ L^{-1} \colon L(X) \to X, y \mapsto x \quad (\forall y = Lx \in L(X)),$

is bounded. Indeed, "formally" we have $ \|x\|_X = \|L^{-1} Lx \|_X \leq c_1 \|L x\|_Z$. Okay, the problem is just that the inverse restricted to $L(X)$ does not need to exist. This happen when the kernel of $L$ contains nonzero elements.

Now you do as the article suggests and you decompose the space $X$ into a direct sum $X = X_1\oplus \operatorname{ker} L$. Then you take arbitrariy $x \in X$ and you write it in a unique way, due to the direct sum, as $x = x_1 + x_2$. What you get is

$\|x\|_X \leq \|x_1\|_X + \|x_2\|_X \leq c_1 \|L x_1\|_Z + c_2 \|x_2\|_Y$.

This is because $\operatorname{ker}L$ is a finite space and any norm taken on this space is equivalent. So you can estimate the norm $\|x_2\|_X$ by $c_2 \|x_2\|_Y$.

Now you use the trick that $Lx_2 = 0$ and you get $c_1 \|L x_1\|_Z = c_1 \|L x_1 + Lx_2\|_Z = c_1 \|L x\|_Z$. And you use that $x_2 = x - x_1$.

Okay this gives you

$\|x\|_X \leq c_1 \|L x\|_Z + c_2 \|x - x_1\|_Y$.

Now you use the injection $X \to Y$ and another time the first argument about the invertibility of $L$ on $X_1$ and get

$\|x\|_X \leq c_1 \|L x\|_Z + c_2 \|x\|_Y + \|x_1\|_Y \leq c_1 \|L x\|_Z + c_2 \|x\|_Y + \|x_1\|_X \leq 2c_1 \|L x\|_Z + c_2 \|x\|_Y$.

Now rename the constant $2c_1$. Okay, the problem is now that we didn't use the closed graph theorem right? Actually that is not a problem because we secretly made use of it. It gave us that $L$ has a bounded inverse on $X_1$. Note that the closed graph theorem is equivalent to the bounded inverse theorem. It states that a bijective bounded linear operator between Banach spaces has a bounded inverse.

2)Summarized version without the explanations: $$\|x\|_X \leq \|x_1\|_X + \|x_2\|_X \\ \leq C_1 \|L x_1\|_Z + c_2 \|x_2\|_Y\\ = C_1 \|L x_1 + L x_2\|_Z + c_2 \|x_2\|_Y \\ = C_1 \|L x \|_Z + c_2 \|x_2\|_Y \\ = C_1 \|L x \|_Z + c_2 \|x - x_1\|_Y \\ \leq C_1 \|L x \|_Z + c_2 \|x\|_Y + \|x_1\|_Y \\ \leq C_1 \|L x \|_Z + c_2 \|x\|_Y + \|x_1\|_X \\ \leq 2C_1 \|L x \|_Z + c_2 \|x\|_Y \\ = c_1 \|L x \|_Z + c_2 \|x\|_Y. $$

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