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](https://mathoverflow.net/questions/359911/existence-of-injective-compact-operators)). 



Existence of an injective unbounded below operator
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.)


