Properties of $||x||_X\leq c||Tx||_Y+||Kx||_Z$ for every $x\in X$ Let $T: X\to Y$ be a bounded linear operator between Banach spaces.Assume that there exists c$\in R$ and a compact operator $K: X\to Z$ such that 
$$||x||_X\leq c||Tx||_Y+||Kx||_Z\quad \forall x\in X$$
Prove that 
a)Ker(T) ha finite dimension
b) if $(x_n)$ is a sequence in X with bounded image and $||x_n||_X\to \infty$ then the sequence $x_n/||x_n||_X$ has a subsequence that converges to an element of Ker(T).
c) T has closed image
Prove that if $T\in L(X,Y)$ has a finite dimension kernel and closed image there exists c and K like before.
I would appreciate any idea on the solution. Thank you in advance.
 A: a) Suppose that $x\in\ker T$. Then $\|x\|_X\leq\|Kx\|_Z$. As $K$ is compact, this can only happen on a finite-dimensional subspace (otherwise, you would be able to put a ball in the image of $K$). 
b) The assumption is that $\|Tx_n\|_Y\leq k$ for all $n$. Let $y_n=x_n/\|x_n\|_X$. Then
$$
\|Ty_n\|_Y=\frac{\|Tx_n\|_Y}{\|x_n\|_X}\leq\frac{k}{\|x_n\|_X}\to0.
$$
The compacity of $K$ allows us to choose a subsequence $y_{n_k}$ with $\{Ky_{n_k}\}$ convergent; in particular, it is Cauchy. As $\|y_{n_k}-y_{n_j}\|_X\leq c\|Ty_{n_k}-Ty_{n_j}\|_Y+\|Ky_{n_k}-Ky_{n_j}\|_Z$, we deduce that $\{y_{n_k}\}$ is Cauchy. As $X$ is Banach, there exists $y\in X$ with $y_{n_k}\to y$. Then, as $T$ is bounded,  $Ty=\lim_kTy_{n_k}=0$, i.e. $y\in\ker T$. 
c) The inequality 
$$
\|x\|_X\leq c\|Tx\|_Y+\|Kx\|_Z
$$
will still hold if we replace $c$ with a smaller positive number. So we can assume that $c\leq\frac1{2\|T\|}$. Then
$$
\|x\|_X\leq c\|Tx\|_Y+\|Kx\|_Z\leq c\|T\|\,\|x\|_X+\|Kx\|_z\leq\frac12\,\|x\|_X+\|Kx\|_Z,
$$
and so 
$$
\|x\|_X\leq 2\|Kx\|_Z
$$
for all $x\in X'$. Let $X_0$ be the closed span of $\{x\in X':\ Kx\ne0\}$. The compacity of $K$ forces $X_0$ to be finite dimensional; it is then complemented in $X'$, and this complement $X_1$ is contained in the kernel of $K$. So $X=\ker T\oplus X_0\oplus X_1$, with the first two finite-dimensional. So it is enough to check that $T(X_1)$ is closed. Let $\{Tx_n\}$ be a Cauchy sequence with $\{x_n\}\subset X_1$. Then $Kx_n=0$ for all $n$ and $\|x_{n_k}-x_{n_j}\|_X\leq c\|Tx_{n_k}-Tx_{n_j}\|_Y$, so $\{x_{n_k}\}$ is Cauchy. As $X_1$ is Banach, there exists $x\in X_1$ with $x_{n_k}\to x$, and then $\lim Tx_n=Tx$. 
d) As $\ker T$ is finite-dimensional, it is complemented in $X$. Let $K$ be the projection onto $\ker T$, i.e. $K$ is the identity on $\ker T$ and zero on its complement $X'$. Now, for any $x\in X$ we have $x=x_1+x_2$, with $x_1\in\ker T$ and $x_2\in X'$.
$$
\|x\|_X=\|x_1+x_2\|_X\leq\|x_1\|_X+\|x_2\|_X=\|Kx_1\|_X+\|x_2\|_X
$$
Now, as $X'$ is closed, it is a Banach space, and the restriction $T:X'\to \text{im}\, T$ is bounded, one-to-one, and onto. So it has a bounded inverse; in particular, $T$ is bounded below on $X'$. So we have
$$
\|x\|_X\leq\|Kx_1\|_X+\|x_2\|_X=\|Kx\|_X+\|x_2\|_X\leq\|Kx\|_X+c\|Tx_2\|_Y=\|Kx\|_X+\|Tx\|_Y.
$$
