Is the image of closed unit ball under the linear operator closed? 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.
 A: 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 $z\in B_1(0)$.
A: I agree with the answer of @thing, but I allow myself to post my own solution, which I reached last night. My solution starts with a similar idea, but more difficult.
Firstly, we take a point $y_0\in Cl(A(B_1^{l_1}(0)))$. So there is a sequence $\{x_n\}\subseteq B^{l_1}_1(0)$ such that $||Ax_n -y_0||_{L_3} \rightarrow 0$ with $n\rightarrow +\infty$. Then we notice that $l_1\subseteq l_{10/9}$. Because for all $n\in \mathbb{N}$ we have that $||x_n||_{10/9}\leq ||x_n||_1\leq 1$, the sequence $\{x_n\}$ is bounded in $l_{10/9}$. Since $l_{10/9}$ is reflexive and separable by the Banach-Tihonov theorem there is a subsequence $\{x_{n_j}\}$, which weakly in the meaning of $l_{10/9}$ converges to some $x_0\in l_{10/9}$.
From weak convergences of $\{x_{n_j}\}$ to $x_0$ we have that for all $k\in \mathbb{N}$ that $x_{n_j}(k)\rightarrow x_0(k)$ with $ j\rightarrow +\infty $. For all $K\in \mathbb{N}$ we have that $\sum\limits_{k=1}^K |x_0(k)|=\lim\limits_{j\rightarrow +\infty} 
\sum\limits_{k=1}^K |x_{n_j}(k)|\leq 1$, so $\sum\limits_{k=1}^{+\infty} |x_0(k)|\leq 1$. It proves that $x_0\in B^{l_1}_1(0)$.
Now the most strange step. I want to prove that $A$ can be considered as a continuous linear operator from $l_{10/9}$ to $L_3(0,+\infty)$. Let $x\in B^{l_{10/9}}_1(0)$. With simple estimations and Helder's inequality it is true that \begin{equation} \int\limits_0^{+\infty} \left|\sum\limits_{k=1}^{+\infty} \frac{x(k)}{\sqrt{k}+\sqrt{t}}\right|^3 dt\leq \int\limits_0^{+\infty} \left(\sum\limits_{k=1}^{+\infty} \frac{|x(k)|}{\sqrt{k}+\sqrt{t}}\right)^3 dt=\sum\limits_{m=0}^{+\infty} \int\limits_{m}^{m+1} \left(\sum\limits_{k=1}^{+\infty} \frac{|x(k)|}{\sqrt{k}+\sqrt{t}}\right)^3 dt\leq\end{equation}
\begin{equation} \leq\sum\limits_{m=0}^{+\infty} \left(\sum\limits_{k=1}^{+\infty} \frac{|x(k)|}{\sqrt{k}+\sqrt{m}}\right)^3 \leq \sum\limits_{m=0}^{+\infty} \left( \left(\sum\limits_{k=1}^{+\infty} |x(k)|^{10/9}\right)^{9/10} \left(\sum\limits_{k=1}^{+\infty} \frac{1}{(\sqrt{k}+\sqrt{m})^{10}}\right)^{1/10} \right)^3.\end{equation}
If $m>0$ we have that $\sum\limits_{k=1}^{+\infty} \frac{1}{(\sqrt{k}+\sqrt{m})^{10}}\leq\sum\limits_{k=1}^{+\infty} \int\limits_{k-1}^k  \frac{1}{(\sqrt{t}+\sqrt{m})^{10}} dt=\int\limits_0^{+\infty} \frac{dt}{(\sqrt{t}+\sqrt{m})^{10}}=\frac{1}{36m^4}$. So we have that \begin{equation} \sum\limits_{m=0}^{+\infty} \left( \left(\sum\limits_{k=1}^{+\infty} |x(k)|^{10/9}\right)^{9/10} \left(\sum\limits_{k=1}^{+\infty} \frac{1}{(\sqrt{k}+\sqrt{m})^{10}}\right)^{1/10} \right)^3 \leq \left(\sum\limits_{k=1}^{+\infty} \frac{1}{k^5} \right)^3+ \left(\sum\limits_{m=1}^{+\infty} \frac{1}{(36m^4)^{3/10}}\right)^3 <+\infty.\end{equation} The last number is finite and does not depend on $x$. So $A$ is bounded linear operator from $l_{10/9}$ to $L_3(0,+\infty)$.
Since $x_{n_j}$ converges weakly in the meaning of $l_{10/9}$ we have for all $f\in (L_3(0,+\infty))^*$ that $\lim\limits_{j\rightarrow +\infty} f(A(x_{n_j}))-f(A(x_0))=\lim\limits_{j\rightarrow +\infty}(f\circ A)(x_{n_j}-x_0)=0$, because $f\circ A$ is linear functional on $l_{10/9}$. So, $Ax_{n_j}$ weakly converges to $Ax_0$.
After all, we have strongly converging to $y_0$ sequence $Ax_n$ and it's weakly converging to $Ax_0$ subsequence $Ax_{n_j}$. That implies that $y_0=Ax_0$.
Because $y_0\in B^{l_1}_1(0)$ we have that $y_0\in A(B^{l_1}_1(0))$. So, $A(B^{l_1}_1(0))$ is closed. Q.E.D.
