Existence of an injective unbounded below operator

Question

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).

Answer 1

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.)