Problem: Show $\phi(x) = 4x_2 - ix_3 + x_6$ , $x \in l^3$ is bounded and determine its norm.
Showing Bounded: $|\phi(x)| \leq |4x_2| + |x_3| + |x_6| \leq 4\sum_{i=1}^{\infty}|x_i| = 4||x||_1$
Question: Is the above true? The step where he bounds by $4||x||_1$ seems false. If true is it also the case that $4||x||_1 \leq c||x||_3$, for some $c \in$ R? I thought the Hölder inequality was applicable, but can't see how to use it here.
For your first question, $$ 4|x_2|+|x_3|+|x_6|\leq 4|x_2|+4|x_3|+4|x_6|\leq4\sum_{j=1}^\infty|x_j|. $$ For your second question: no, there is no $c$ with $\|x\|_1\leq c\|x\|_3$ for all $x$.
What they should have done is to use such a bound while they had finitely many terms. Namely, for $a\geq b,\geq c\geq0$ you have the estimate $$ (a+b+c)^3\leq (3a)^3=27a^3\leq 3^3(a^3+b^3+c^3). $$ So, $$ 4(|x_2|+|x_3|+|x_4|)\leq 12 (|x_2|^3+|x_3|^3+|x_4|^3)^{1/3}\leq 12\left(\sum_j|x_j|^3\right)^{1/3}=12\|x\|_3. $$