Binding compact operator between Banach spaces

Question

For a graded hand-in, we have to solve the following problem. I think I have a solution, but it seems a bit too easy and also illogical given the proposition, so I would like to get a hint of where I go wrong.

The problem:

Let X, Y, Z be Banach spaces, take $K \in K(X,Y)$, where $K(X,Y)$ denotes the set of compact operators from X to Y, and take $J \in B(Y,Z)$, J one-to-one, where $B(Y,Z)$ denotes the set of bounded operators from Y to Z. Show that for every $\epsilon > 0$, there exists a constant $C_\epsilon > 0$ st.:

$ ||Kx|| \leq \epsilon ||x|| + C_\epsilon ||JKx||, \quad x \in X.$.

My solution:

$||Kx|| = ||J^{-1}JKx|| \leq ||J^{-1}||*||JKx|| \leq \epsilon||x|| + ||J^{-1}||*||JKx||$

By the bounded inverse theorem, $J^{-1} \in B(R(J)\subset Z,Y)$, so $||J^{-1}|| \in (0,\infty)$.

There are bounded operators $J : Y\rightarrow Z$ that are one-to-one, but with unbounded inverses. — Disintegrating By Parts, Oct 31 '16 at 03:47
Thank you, you're right, I cannot assume that R(J) is closed and thus a Banach space — Scipio, Oct 31 '16 at 09:34

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

As TrialAndError noted, R(J) need not be a Banach space and therefore the bounded inverse theorem does not apply. See this question for further discussion: When is the image of a linear operator closed?.

edited Apr 13 '17 at 12:20

Community

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answered Oct 31 '16 at 09:36

Scipio

390

Binding compact operator between Banach spaces

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