Show that the linear functional $f(x) = 3\xi_1 - 5\xi_2$ is bounded on $\ell^1$ and compute its norm How do I show that $f$ is bounded and determine the norm of $f$?
Starting with,
$|f(x)| = |3\xi_1 - 5\xi_2|$ 
I am not sure what the next move is to get to the form
$|f(x)| \leq c\|x\|$ where $\|x\| = \sum_{k=1}^{\infty} |\xi_k|$ 
 A: I assume $x=(\xi_i)_i$ ?
Hints: 
1) For all $a,b\in\mathbb C$ there holds $|a\pm b|\leq|a|+|b|$ and $|ab|=|a|\cdot|b|$. So, $|3\xi_1-5\xi_2|\leq \ldots$
2) $|\xi_1|\leq \|x\|$ and $|\xi_2|\leq \|x\|$
A: If the sequence
$(x) = (\xi_i)_{i = 0}^\infty = (\xi_1, \xi_2, \ldots, \xi_n, \ldots ), \tag 1$
with
$\Vert (x) \Vert = \displaystyle \sum_1^\infty \vert \xi_i \vert < \infty, \tag 2$
then with
$f(x) = 3\xi_1 - 5 \xi_2, \tag 3$
we have
$\vert f(x) \vert = \vert  3\xi_1 - 5 \xi_2 \vert \le \vert 3\xi_1 \vert + \vert 5 \xi_2 \vert = 3\vert \xi_1 \vert + 5 \vert \xi_2 \vert$
$\le 5(\vert \xi_1 \vert + \vert \xi_2 \vert) \le 5 \displaystyle \sum_1^\infty \vert \xi_i \vert = 5 \Vert (x) \Vert, \tag 4$
which shows
$\Vert f \Vert \le 5; \tag 5$
we can in fact show that
$\Vert f \Vert = 5 \tag 6$
by taking
$(x) = (0, 1, 0, 0, \ldots, 0, \ldots ); \tag 7$
then
$\Vert (x) \Vert = 1, \tag 8$
and
$\vert f(x) \vert = 5 = 5 \Vert (x) \Vert, \tag 9$
which implies the smallest possible real $c > 0$ such that 
$\vert f(x) \vert \le c \Vert 5 \Vert \tag{10}$
for all $(x)$ must obey
$c \ge 5; \tag{11}$
this combined with (5) yields
$\Vert f \Vert = 5. \tag{12}$
A: By definition
$$\|f\|:=\sup\left\{|3\xi_1-5\xi_2|\ \biggm|\ \sum_{i=1}^\infty|\xi_i|=1\right\}\ .$$It is then pretty clear that in fact
$$\|f\|=\sup\bigl\{|3\xi_1-5\xi_2|\>\bigm|\>|\xi_1|+|\xi_2|\leq1\bigr\}\ .$$
Now $|\xi_1|+|\xi_2|\leq1$ implies
$$|3\xi_1-5\xi_2|\leq 3|\xi_1|+5|\xi_2|\leq5\bigl(|\xi_1|+|\xi_2|\bigr)\leq5\ ,$$
which allows to conclude that $\|f\|\leq 5$. On the other hand we have $\bigl|f(0,1,0,0,\ldots)\bigr|=5$, so that we obtain $\|f\|= 5$.
A: The dual space of $\ell^1$ is isometrically isomorphic to the space $\ell^\infty$. Furthermore, every bounded linear map $f : \ell^1 \to \mathbb{R}$ can be expressed as
$$f(x_n)_{n=1}^\infty = \sum_{n=1}^\infty \alpha_nx_n$$
for some sequence $(\alpha_n)_{n=1}^\infty \in \ell^\infty$, and vice-versa, every linear map of this form is in $(\ell^1)^*$. The isometrical isomorphism is given by $f \mapsto (\alpha_n)_{n=1}^\infty$, so in particular $\|f\| = \left\|(\alpha_n)_{n=1}^\infty\right\|_\infty$.
For your functional in particular we have that $f$ is represented by the vector $(3, -5, 0, 0, \ldots) \in \ell^\infty$ so we conclude that $f$ is bounded and $$\|f\| = \|(3, -5, 0, 0, \ldots)\|_\infty = \max\{|3|, |-5|\} = 5$$ 
