I am just starting functional analysis and we were asked two questions by our teacher.
Here's the first question:
Let $l^{\infty}(\mathbb{N})$ be the vector space of functions, $x : \mathbb{N} \rightarrow \mathbb{R}$ such that $$ \sup_{n \in \mathbb{N}} \left| x(n) \right| < \infty.$$
On $l^{\infty}(\mathbb{N})$, let us now consider the following two norms: $$\left\Vert x \right\Vert_{\infty} = \sup_{n \in \mathbb{N}} \left| x(n) \right|,\ \ \ \ \ \ \left\Vert x \right\Vert _* = \sum_{n \in \mathbb{N} } 2^{-n} \left| x(n) \right|.$$
Let $\{ \alpha_k \}_{k=1}^{\infty}$ be a bounded sequence of positive real numbers such that $\alpha := \inf_{n\in \mathbb{N}} \alpha_n > 0$.
Show that the linear transformation $$T:(l^{\infty}(\mathbb{N}),\left\Vert \ \cdot \ \right\Vert_*) \rightarrow (l^{\infty}(\mathbb{N}),\left\Vert \ \cdot\ \right\Vert_{\infty})$$ defined by $Tx(k) = x(k) \alpha(k) (\forall k \in \mathbb{N})$, is not continuous.
And here's the second question:
Let $l^1(\mathbb{N})$ be the vector space of functions, $x : \mathbb{N} \rightarrow \mathbb{R}$ such that $$ \left\Vert x \right\Vert_1 = \sum_{n \in \mathbb{N}} | x(n) | < \infty.$$
Let $\{ \alpha_k \}_{k \in \mathbb{N}}$ be a bounded sequence of positive real numbers and let $m = \sup_{k \in \mathbb{N}} |\alpha _k|$. Define $$T:(l^1(\mathbb{N}),\left\Vert \ \cdot \ \right\Vert_1) \rightarrow (l^{\infty}(\mathbb{N}),\left\Vert \ \cdot \ \right\Vert_{\infty}),$$ by $$Tx(n) = \sum_{k = 1} x(k) \alpha(k) (\forall n \in \mathbb{N}).$$
Show that $\left\Vert T \right\Vert = m$.
Any help on both of these will be greatly appreciated! Thanks!
For $a)$ consider $e_n$ the sequence such that $e_n(i)=0, i\neq n, e_n(n)=2^n$, $\|e_n\|_*=1$ and $T(e_n)(i)=0, i\neq n, T(e_n)(n)=\alpha(n)2^n>\alpha 2^n$ implies that $\|T(e_n)\|_\infty$ is not bounded.
For the second show that $T$ is bounded and its norm is inferior to $m$ and consider $e_n$ such that $lim_n\alpha_n=m$ to show that $\|T\|=m$.