# Existence of an injective unbounded below operator

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).

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.)