Let $B_nx=x(t-\frac{1}{n})$ be a sequence of operators in $L^2(\mathbb{R})$. Show that $B_n\overset{s}{\to}Id$ but $B_n\not\rightrightarrows Id$ where $Id$ is the identity operator.
So for the strong convergence part I said $||B_nx-Idx||=||x(t-\frac{1}{n})-x(t)||\to0$ as $n\to\infty$ because $x(t-\frac{1}{n})\to x(t)$ as $n\to\infty$ (not quite sure if this is sufficient)
On the other hand, I am not sure how to show that uniform convergence is not true. I am trying to maybe show that $||B_n-Id||$ is equal to some constant. I did the following: $$||B_nx-Idx||^2=\int_{-\infty}^\infty(x(t-\frac{1}{n})-x(t))^2dx=\int_{-\infty}^\infty x^2(t-\frac{1}{n})dx-2\int_{-\infty}^\infty x(t-\frac{1}{n})x(t)dx+\int_{-\infty}^\infty x^2(t)dx$$ $$\leq\int_{-\infty}^\infty x^2(t-\frac{1}{n})dx-2||Idx||||B_nx||+\int_{-\infty}^\infty x^2(t)dx\leq||Idx||^2+||B_nx||^2=2||x||^2$$ So $||B_n-Id||\leq\sqrt{2}$ (I am assuming this part is right)
Then can I can find some function where this $\sqrt{2}$ is achieved so I can conclude that $||B_n-Id||=\sqrt{2}$ and $B_n\not\rightrightarrows Id$?
Hints for strong convergence:
Using the Lebesgue dominant convergence theorem prove that if $x(t)$ is compactly supported and continuous, then $\|B_nx(t)-x(t)\|\to0$ (if support of $x(t)$ is $[a,b]$, then $|B_nx(t)-x(t)|\leq\max\limits_{t\in[a,b]}|x(t)|(\chi_{[a,b]}(t)+\chi_{[a,b+1]}(t))$);
Using the density of continuous compactly supported functions and the Banach-Steinhaus theorem, prove strong convergence
Hint for failure of uniformly convergence: consider the sequence $x_n(t)=\sqrt n\chi_{[0,\frac{1}{n})}(t)$.