# 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.

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 $$z\in B_1(0)$$.

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.