Let $A:c_0\to L_2[0,1]$, $Ax(t)=\displaystyle\sum\limits_{k=1}^\infty\dfrac{x(k)}{ k!+ t}$ — is linear operator. It is required to prove or disprove the closedness of the set $A(B_1(0))$, where $B_1(0)$ — closed unit ball in $c_0$.
What I already know about this operator: it's well defined, bounded, compact. I can find his adjoint. If instead of $c_0$ there was a reflexive space, I could say that the set $A(B_1(0))$ is closed (even compact). Is it possible to use the above information for this case, or should I act somehow differently?
Best Answer
Hint:
Consider the function $$ f(t)=\sum_{k=1}^\infty \frac{1}{k!+t}, \qquad t\in[0,1]. $$ To show this is not in $A(B_1(0))$, one can use monotonicity arguments. For example, it turns out that $(Ax)(t)\leq f(t)$ for all $t\in[0,1]$ and $x\in B_1(0)$ holds. Similar ideas can be used to show that $f\not\in A(B_1(0))$.