Injective continuous operators between Banach spaces

Of course the dimension is an obvious obstacle, but even if the space have the same cardinality of Hamel bases the answer is no. For example in the paper

A. Avilés, P. Koszmider, A Banach space in which every injective operator is surjective. Bull. Lond. Math. Soc. 45 (2013), no. 5, 1065–1074

the authors constructed an infinitely dimensional Banach space $X$ such that if $T:X\to X$ is bounded and injective, then $T(X)=X$. Therefore if $Y$ is a subspace of $X$, then one cannot find an injective operator $T:X\to Y$.

Piotr Hajłasz' answer nails the problem, however, let me point out that there are easier examples of such pairs of spaces among spaces that have the same density.

Suppose that $X$ fails to have a strictly convex renorming. Thus, there is no injective operator $T$ from $X$ into any space $Y$ that is strictly convex, as if it were $\|x\|^\prime = \|x\| + \|Tx\|$ would be a strictly convex norm on $X$.

Spaces that do not have a strictly convex norm include

• $X = \ell_\infty^c(\Gamma)$, the space of all bounded scalar-valued functions on an uncountable set $\Gamma$ that have at most countable support (Day);
• $X = \ell_\infty / c_0$ (Bourgain).

In the latter case, you may even take $Y= \ell_\infty$.