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Interesting inequality involving Banach spaces and an operator


Let E,F be two Banach spaces with norms \|\cdot \|_E, \ \|\cdot \|_F. Let T \in \mathcal{L}(E,F)(space of linear bounded operators T:E \to F) be such that R(T) is closed and \dim N(T)< \infty. Let | \cdot | denote another norm on E which is weaker than \|\cdot \|_E, i.e. |x | \leq M \|x\|_E, \ \forall x \in E.

Prove that there exists a constant C such that \|x\|_E \leq C(\|Tx\|_F+|x|),\ \forall x \in E.

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Chapter 2

Proof: Argue by contradiction. Suppose there exists a sequence of elements x_n \in E with \|x_n\|=1 and \|Tx_n\|+|x_n| < \frac{1}{n},\ \forall n\geq 1. We can assume that T is surjective; otherwise, replace F by R(T). Then by the open mapping theorem, there exists c>0 such that T B_E(0,1) \supset c B_F(0,1). Therefore, there exists y_n \in E,\ \|y_n\|< \frac{1}{nc} such that Tx_n=Ty_n. Therefore x_n=y_n+z_n with z_n \in N(T) and \|y_n\| \to 0, \ \limsup \|z_n\| \geq 1. But N(T) is finite dimensional, and with |\cdot | it becomes a Banach space. On a Banach space two norms that are comparable are equivalent, and therefore \|\cdot \|_E,\ |\cdot | are equivalent. We have that |z_n| \leq \frac{1}{n}+M \|y_n\|_E \to 0, which contradicts the fact that \limsup \|z_n\|_E \geq 1. Therefore, the assumption made is false, and the problem is solved.

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