$\ell_2$ convergence and $\ell_1$ norm convergence implies $\ell_1$ convergence

Question

Let $x_n \in \ell_2$ converge to $x_\infty \in \ell_2$ and $||x_n||_1$ converge to $||x_\infty||_1$ where $||\cdot||_1$ is $\ell_1$ norm. Is it true, that $x_n$ converge to $x_\infty$ in $\ell_1$?

Answer 1

No, if you allow $||x_\infty||_1=\infty$: let $$x_n(j)= \begin{cases} 1/j,&(1\le j\le n), \\0,&(j>n).\end{cases}$$

Yes if $||x_\infty||_1<\infty$. Let $\epsilon>0$. Choose $A$ so $$\sum_{j=A+1}^\infty|x_\infty(j)|<\epsilon.$$Since convergence in $\ell^2$ implies pointwise convergence there exists $N$ so that $$\left|\vert \vert x_n\vert\vert_1-\vert\vert x_\infty\vert\vert_1\right|<\epsilon\quad(n>N)$$and $$\sum_{j=1}^A|x_n(j)-x_\infty(j)|<\epsilon\quad(n>N).$$The triangle inequality implies $$\sum_{j=1}^A|x_n(j)|\ge\left(\sum_{j=1}^A|x_\infty(j)|\right)-\epsilon\ge||x_\infty||_1-2\epsilon\quad(n>N),$$hence $$\sum_{j=A+1}^\infty|x_n(j)|<3\epsilon\quad(n>N),$$which shows finally that $$||x_n-x_\infty||_1<5\epsilon\quad(n>N).$$

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In fact the same holds for any measure space. Say $\mu$ is a measure on $X$, $||f_n-f||_2\to0$, $||f||_1<\infty$ and $||f_n||_1\to||f||_1$. Start by choosing a set $E$ with $\mu(E)<\infty$ and $$\int_{X\setminus E}|f|\,d\mu<\epsilon,$$and use the fact that $$\int_E|f_n-f|\,d\mu\le(\mu(E))^{1/2}||f_n-f||_2.$$