Let $T:\ell^2\to \ell^2$ defined by $(x_1,x_2,x_3,x_4,\dots) \mapsto (0,4x_1,x_2,4x_3,x_4,\dots)$. Prove $T$ is continuous.
My attempt: $T$ is well defined because for $x\in\ell^2$ we have: $\sum_{n=1}^\infty |x_n|^2 <\infty$ implies $\sum_{n=1}^\infty |x_{2n}|^2 + \sum_{n=1}^\infty |x_{2n-1}|^2<\infty$ implies $\sum_{n=1}^\infty |x_{2n}|^2 + 4\sum_{n=1}^\infty |x_{2n-1}|^2<\infty$. So $T(x) \in\ell^2$ for all $x\in\ell^2$. To prove continuity I need to find a scalar $k>0$ such that $\parallel T(x)\parallel\leq k \parallel x\parallel$. I tried: Let $y$ be all even elements of $x$, $z$ all uneven elements.
$$\parallel T(x)\parallel_2 = \parallel (0,4x_1,x_2,4x_3,x_4,\dots)\parallel_2 \leq \parallel y\parallel_2 + \parallel 4z\parallel_2 = \parallel y\parallel_2 + 4\parallel z\parallel_2$$ Now I wonder is $k=4$ is the lowest possible value of $k$.
Because I also tried to compute $\parallel T\parallel:= \sup\{\parallel T(x)\parallel_2 : \parallel x\parallel_2\leq 1\}$. Which gave me \parallel T\parallel_2 = 4. But now the thing is, that I computed $\parallel T^2\parallel = 4$. How does his work?
You have that $\Vert T(x)\Vert_2\le 4\Vert x\Vert_2$, so $\Vert T\Vert \le 4$. If you take $x=y+z$ such that $y=0$ and $\Vert z\Vert_2=1$ you get $\Vert T(x)\Vert_2=4$, so $\Vert T\Vert \ge4$. The best constant is 4. In general you only have $\Vert T^2\Vert \le \Vert T\Vert^2 $ (see enter link description here) so there is no contradiction