Definition of a compact linear operator

Question

We know that if $X$ and $Y$ are Banach spaces, then a linear $T:X\to Y$ is compact if $T(B_X)$ is relatively compact, i.e. $\overline{T(B_X)}$ is compact. My question is why it is not defined as $T(B_X)$ is compact? Does this imply $Y$ is finite dimensional? Any comment in this regard will be appreciated.

Answer 1

If $B_X$ is the closed unit ball (which is the standard notation) then it is possible for $T(B_X)$ itself to be compact, contrary to some of the claims made in the previous comments. In fact, if $X$ is reflexive that this is always true. Let $\|x_n\| \leq 1$ and $Tx_n \to y$. Then there is a subnet $\{x_{n_{k}}\}$ which converges weakly, say to $z$. The $\{T(x_{n_{k}})\} \to Tz$ weakly which implies $y=Tz$. Thus $T(B_X)$ is closed. Thus the new definition you are suggesting is equivalent to the usual definition in reflexive spaces. A counterexample in the general case is the following: let $Tf(x)=\int_0^{x}f(t) \, dt$ for $f\in C[0,1]$. The $T$ is compact but the image of the closed unit ball is not closed. [Take a nondiferentiable function and approximate it by polynomials. These polynomials are in $T(B_X)$ but their limit is not].