4
$\begingroup$

Given an infinite dimensional Banach space, can we always find an infinite dimensional Banach space $Y$ and an injective bounded operator $T:X\to Y$ such that $T$ is not bounded below?

If $X^{*}$ is $w^*$-separable, then for every Banach space $Y$, there exists an injective compact operator $T: X\to Y$ (see Goldberg and A.H. Kruse, The Existence of Compact Linear Maps Between Banach Spaces. Proc. A.M.S. 13 (1962), 808-811) and we know that a compact operator is not bounded below.

We also know that for $X$ nonseparable we may not find any injective compact operator into any Banach space (Existence of injective compact operators).

$\endgroup$
4
$\begingroup$

Yes. Using the injective property of $\ell_\infty$, get an operator $S:X\to \ell_\infty$ that is compact on some infinite dimensional subspace $X_0$ of $X$. Let $Q: X \to X/X_0$ be the quotient map. Define $T:X\to \ell_\infty \oplus X/X_0$ by $Tx = (Sx, Qx)$.

With a slightly different argument you can replace $\ell_\infty$ by any infinite dimensional Banach space with $S$ being even nuclear.

A more challenging problem is whether there is always an injective strictly singular operator from $X$ into some Banach space. (Hint: The answer is negative.)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.