Question about operators on Banach spaces with closed range and finite-dimensional kernel Let $A,B$ be two complete normed linear spaces with norms $||\cdot||_A ,||\cdot||_B$. Let $S \in \mathcal{L}(A,B)$ and assume its range is closed and kernel is finite-dimensional. Let $||\cdot||$ be another norm on A such that $||x|| \leq C||x||_A, \forall x \in A$ for a constant $C > 0$. Prove that there exists a constant $D > 0$ such that $||x||_A \leq D(||Sx||_B + ||x||), \forall x \in A$.
A hint is given to use the argument by contradiction. However, I still have no idea how I can solve this after thinking about it for a long time. The tools I can think of that might be useful is the closed graph theorem and the result that says if $A,B$ are Banach and $S \in \mathcal{L}(A,B)$ is bijective, then $S^{-1} \in \mathcal(B,A).$ Also, I don't know how to use the condition that the kernel is finite-dimensional. The only thing I can deduct from that is $A/\text{Ker}S$ is complete, and I also know the range is complete. But again, I have no idea how to proceed with these...
Any help is greatly appreciated! Thank you!
 A: Make use of the following facts:


*

*On finite dimensional spaces all norms are equivalent. Hence there is some $k$ so that $\|x\|_A ≤ k\|x\|$ for all $x\in \ker(S)$.

*Since $\ker(S)$ is finite dimensional there is a continuous projection $P:A\to A$ onto $\ker(S)$. $Y:=\ker(P)$ is then a complement of $\ker(S)$ and also a Banach space.


Use this complement to see that for $x\in A$:
\begin{align}
\|x\|_A &=\|Px+(1-P)x\|_A≤ \|Px\|_A+\|(1-P)x\|_A ≤ k\,\|Px\|+\|(1-P)x\|_A\\
&≤k\,\|Px+(1-P)x-(1-P)x\|+\|(1-P)x\|_A≤ k\,\|x\|+k\,\|(1-P)x\|+\|(1-P)x\|_A\\
& ≤ k\,\|x\|+(1+kC)\|(1-P)x\|_A.
\end{align}
Now $(1-P)x$ is in $Y$. The map $S:Y\to\mathrm{im}(S)$ is continuous, injective, surjective and has codomain a Banach space (because we know that $\mathrm{im}(S)$ is closed). Hence it is invertible. This means that there is a constant $K2$ with $\|(1-P)x\|_A ≤ K\, \|S(1-P)x\|_B$. Note that $S(1-P)=S$, hence $\|(1-P)x\|≤K\,\|Sx\|_B$. Combine both inequalities to get:
$$\|x\|_A ≤ k\,\|x\|+(1+kC)K\,\|Sx\|_B.$$
A: By contradiction, suppose that for each $n\in \mathbb{N}$, there is $x_n$ in $A$ such that $$\frac{1}{n}||x_n||_A > ||Tx_n||_B + ||x_n||$$
Without loss of generality, we may assume that $||x_n||_A = 1$, so that
$$\frac{1}{n} > ||Tx_n||_B + ||x_n||$$
Hints/notes:
1) Without loss of generality, we may assume T is surjective (otherwise replace B with Ran(T))
2) By the open mapping theorem, there is a constant c such that $T(B_A) \supset cB_B$
3) Using 2, write $x_n = y_n + z_n$, where $z_n \in ker(T)$, and $y_n \to 0$
4) Derive a contradiction with the $z_n$ sequence by using the fact that we have equivalent norms on $ker(T)$.
