# Banach space closed graph theorem estimate

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.

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: $$\|x\|_X \leq \|x_1\|_X + \|x_2\|_X \\ \leq C_1 \|L x_1\|_Z + c_2 \|x_2\|_Y\\ = C_1 \|L x_1 + L x_2\|_Z + c_2 \|x_2\|_Y \\ = C_1 \|L x \|_Z + c_2 \|x_2\|_Y \\ = C_1 \|L x \|_Z + c_2 \|x - x_1\|_Y \\ \leq C_1 \|L x \|_Z + c_2 \|x\|_Y + \|x_1\|_Y \\ \leq C_1 \|L x \|_Z + c_2 \|x\|_Y + \|x_1\|_X \\ \leq 2C_1 \|L x \|_Z + c_2 \|x\|_Y \\ = c_1 \|L x \|_Z + c_2 \|x\|_Y.$$