Is the image of closed unit ball under the linear operator closed

Question

I have the linear operator $A:l_1 \rightarrow L_3(0,+\infty)$ which is defined with the formula $(Ax)(t)=\sum\limits_{k=1}^{+\infty} \frac{x(k)}{\sqrt{k}+\sqrt{t}} $ for all $t\in (0,+\infty)$ and for all $x\in l_1$. The full task sounds like this: find the adjoint operator $A^*$ for the operator $A$, then check if the set $A(B_1(0))$ is totally bounded and closed in $L_3(0,+\infty)$. I have found the adjoint operator, then using the criterion of Kolmogorov and Frechet I proved total boundedness of $A(B_1(0))$. But I do not know how to check this set for closedness. It seems difficult to check it with the definition. Also, I cannot see what theorems I may apply. This task is given in the topic about adjoint operators, so maybe I should use some results about adjoint operators, but I am not sure about it.

Answer 1

Consider the sequence $x_n\in B_1(0)\subset l_1$ and let $Ax_n\to y$ in $L_3(0,\infty)$. Then $\exists Ax_{n_k}\to y$ almost everywhere, i.e. $\displaystyle\sum\limits_{s=1}^{\infty} \frac{x_{n_k}(s)}{\sqrt{s}+\sqrt{t}}\to y(t)$ a.e. As you know, a closed unit ball in $l_1$ lies in a closed unit ball of space $l_2$. Since $l_2$ is reflexive, $\exists x_{n_{k_r}}\rightharpoonup z\in l_2$ and $\|z\|_2\leq1$ (since the closed unit ball in $l_2$ is weakly sequentially compact). Since $\forall N\in\mathbb{N}$ $\displaystyle\sum\limits_{s=1}^{N}|x_{n_{k_r}}(s)|\leq1$ and weak convergence implies the coordinatewise, then $\forall N\in\mathbb{N}$ $\displaystyle\sum\limits_{s=1}^{N}|z(s)|\leq1$, so $\|z\|_1\leq1$. Next, for $\varepsilon>0$ choose $M$ so, that $\dfrac{2}{\sqrt{M+1}}<\dfrac{\varepsilon}{2}$ and consider $u=\left(\dfrac{1}{\sqrt1+\sqrt t},\dfrac{1}{\sqrt2+\sqrt t},...,\dfrac{1}{\sqrt M+\sqrt t},0,0,...\right)\in l_2$, then $$\left|\sum\limits_{s=1}^{\infty} \frac{x_{n_{k_r}}(s)}{\sqrt{s}+\sqrt{t}}-\sum\limits_{s=1}^{\infty} \frac{z(s)} {\sqrt{s}+\sqrt{t}}\right|\leq\left|\sum\limits_{s=1}^{M} \frac{x_{n_{k_r}}(s)-z(s)}{\sqrt{s}+\sqrt{t}}\right|+\left|\sum\limits_{s=M+1}^{\infty} \frac{x_{n_{k_r}}(s)-z(s)}{\sqrt{s}+\sqrt{t}}\right|\leq$$$$\leq\left|\sum\limits_{s=1}^{\infty} (x_{n_{k_r}}(s)-z(s))u(s)\right|+\dfrac{2}{\sqrt{M+1}}<\varepsilon,$$ for large enough $r$, because first sum tends to $0$.

Thus, we have $Ax_{n_{k_r}}=\displaystyle\sum\limits_{s=1}^{\infty} \frac{x_{n_{k_r}}(s)}{\sqrt{s}+\sqrt{t}}\to\sum\limits_{s=1}^{\infty} \frac{z(s)} {\sqrt{s}+\sqrt{t}}=Az=y$ a.e., where $z\in B_1(0)$.