Is this linear functional bounded? Find the norm. $$\ell^2\ni (x_n)\rightarrow2x_{1}+28x_2+35 x_{3}$$

I think it can be bounded:

$$|2x_{1}+28x_2+35 x_{3}| \le |2x_{1}|+|28x_2|+|35 x_{3}| \le 65 (\sum_{n=0}^{\infty}|x_n|^2)^{1/2}$$

But I can't find norm.
What is the norm of the linear functional $L(x)$? It is the smallest constant $M$ such that
$$|L(x)| \leq M ||x||_2$$
holds for every $x \in \ell^2$.

Okay, so now note that your linear functional is of the form $\langle a, x \rangle$, so apply the Cauchy inequality to get

$$|L(x)| = |\langle a, x \rangle| \leq ||a||_2 ||x||_2.$$

This means $M \leq ||a||_2$. Now use the fact that $a_1 x_1 + a_2 x_2 + a_3 x_3 = ||a||_2 ||x||_2 \cos \theta$, provided $x_n = 0$ for $n&gt;3$, to show that if $M &lt; ||a||_2$, you could pick an $\theta$ and hence an $x$, such that $|\langle a, x \rangle| &gt; M ||a||_2 ||x||_2$.
Because the dual space of $l^2$ is $l^2$.