# Existence of injective compact operators

We know that if $$X$$ is a separable Banach space, then for every infinite dimensional Banach space $$Y$$, there exists an injective compact operator from $$X$$ to $$Y$$.

My query is for every Banach space $$X$$ (need not be separable ) do there exist a Banach space $$Y$$ and an injective compact operator $$T:X\to Y$$?

• Yes. I have corrected it. – Anupam May 10 at 12:11

No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $$X$$ to have bigger cardinality, for example, $$X = \ell_\infty^*$$. Then you have no chance of building such operators.
• I was actually trying to check that it never exists if $X$ is non-separable. – YCor May 10 at 12:16
• @YCor, take $T\colon \ell_\infty\to c_0$ given by $T (\xi_k)_{k=1}^\infty = (\xi_k / k)_{k=1}^\infty$. It is compact and injective. – Tomasz Kania May 10 at 12:17
• Perhaps it is worth pointing out that $X$ has the property if and only if $X^*$ is weak$^*$ separable, which is a somewhat weaker condition than that $X$ embeds isomorphically into $\ell_\infty$. – Bill Johnson May 10 at 13:59
• @BillJohnson, with the first known example being your space $J\! L$ :-) – Tomasz Kania May 10 at 15:39