inequality of norm for compact operators on Banach space

Question

I have to prove the following statement:

Let $(X,||\cdot||_X),(Y,||\cdot||_Y),(Z,||\cdot||_Z) $ Banach spaces, let $T \in L(X,Y)$ be compact and let $J \in L(Y,Z)$ be injective. Prove that for every $\epsilon \in (0,\infty)$, there exists $C\in [0,\infty)$ such that $$||Tx||_Y\leq \epsilon ||x||_X + C||JTx||_Z \quad \forall x\in X$$

My proof:

Assume the opposite in order to derive a contradiction: I.e.$\exists \epsilon >0, x\in X $ such that $\forall C>0 \quad ||Tx||_Y> \epsilon||x||_X + C||JTx||_Z$
It is easy to see $x$ can not be an element of the kernel of $T$. Hence assume $x\notin \ker(T)$ Since the above inequality holds for abitrarily large $C$ it holds $||JTx||_Z = 0$ otherwise pick $C> \frac{| ||Tx||_Y-\epsilon||x||_X|}{||JTx||_Z}$ But since $J$ is injective $\ker J = \{0\}$ hence $Tx = 0$ which is a contradiction to $x\notin \ker T$

My Question: Is this proof correct? I am asking because I did not require the compactness of $T$. So where does it fail?

Answer 1

Assume $\|x_n\|=1$ and $$\|T\|\ge \|Tx_n\|\ge \varepsilon +n\|JTx_n\|$$ By compactness the sequence $Tx_n$ contains a convergent subsequence $Tx_{n_k}.$ Let $y=\lim_k Tx_{n_k}.$ Then $JTx_{n_k}\to Jy.$ If $Jy\neq 0,$ then $n_k\|JTx_{n_k}\|$ tends to $\infty.$ Therefore $Jy=0,$ i.e. $y=0.$ On the other hand $\|Tx_{n_k}\|\ge \varepsilon,$ which implies $\|y\|\ge\varepsilon,$ a contradiction.

Answer 2

You negated the statement incorrectly. The correct negation is "there exists an $\varepsilon > 0$ such that for all $C > 0$ there exists an $x > 0$..." which means that as you vary $C$ in your argument, you must vary $x$ also.

In particular, as $C$ gets large, it may be that $\|JTx_C\|$ gets small.