functional analysis - Norm of an $l_p$ -operator

Let $F:l_1\rightarrow l_1$ with $F(a_1,a_2,...)=((1-\frac{1}{1})a_1,(1-\frac{1}{2})a_2,(1-\frac{1}{3})a_3,...)$.

a. How do I show that $F$ is bounded?
b. How do I show that $||F||=1$?
c. Is there an $a\in l_1$ such that $||a||=1$ and $|F(a)|=||F||$?

Things I know so far:
a Boundedness has the following definition: that there exists a constant $K$ such that $||F(a)||\leq K||a||$ for all $a\in l_1$. Further the norm is $||a||=\sum_{i=1}^\infty|a_i|<\infty$.
This means that we should have $||F(a)||=\sum_{i=1}^\infty|(1-\frac{1}{i})a_i|\leq K \sum_{i=1}^\infty|a_i|$. Which constant $K$ do we need?

b The norm of an operator $F$ is $||F||=\sup\{|F(a)|:||a||\leq1\}$ and this must equal 1. How can I show this?

c My intuition tells me there is, but I can't think of a way of making it concrete.

Answer

a)

$||F(a)||=\sum_{i=1}^\infty|(1-\frac{1}{i})a_i|\leq \sum_{i=1}^\infty|a_i|$,

since $ 0 \le |1-\frac{1}{i}| \le 1$. Thus take $K=1$ and s0 $||F|| \le 1$

b)

Let $e_n:=(0,....,0,1,0,,,,)$ (1 on the n-th place). Show that $F(e_n)=(1-\frac{1}{n})e_n$.

Thus $1-\frac{1}{n} =||F(e_n)|| \le ||F|| \le 1$ for all $n$. This gives $||F||=1$

c)

Now its your turn to show that there is no(!) $a\in l_1$ such that $||a||=1$ and $||F(a)||=1$