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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.