Bounded functional and extension of it Let $1 ≤ p < ∞,$ and let $H$ be a subspace of $\left(l_p(\Bbb N),||· ||_p \right). $
$$H =\left\{(x,y,z,0,0,...) : x,y,z \in \Bbb K\right\}.$$ 
Let $f : H\to \Bbb K$ be given by $f(x,y,z,0,0,...) = x−y−z$, for all $x,y,z \in\Bbb K$. 
(a) Show that $f$ is bounded on $\left(H,||· ||_p\right)$ and determine $|| f ||$, for each  $1 ≤ p < \infty $. 
(b) Show that if $1 < p < ∞$, then there exist a unique linear functional $F$ on $l_p(\Bbb N)$ extending $f$ and satisfying $||F || = ||f ||$ and show that if $p = 1$, then there are inﬁnitely many linear functional F on $l_1(N)$ extending f and satisfying $||F || = ||f ||$.

For (a) I thought 
 $|f(x)| = |x-y-z| \leq |x|+|y|+|z| = ||x + y+ z||_1  \leq ||x + y+ z||_p \leq \left(|x|^p+|y|^p+|z|^p\right)^{\frac{1}{p}}\leq C||x ||_p$
and for (b)
maybe Hahn-Banach theorem could be used for existence.
 A: The correct estimation is
$$
|f(v)|\leq |x|+|y|+|z|\leq (|x|^p+|y|^p+|z|^p)^{\frac{1}{p}}(1+1+1)^{1 -\frac{1}{p}} =3^{1 -\frac{1}{p}}\lVert v\rVert_p,
$$for $v = (x,y,z,0,0,\ldots)$ and $p\in (1,\infty)$. Since equality can be attained for $v=(1,-1,-1,0,\ldots)$ we have $$\lVert f\rVert_p = 3^{1 -\frac{1}{p}}$$ for $p\in (1,\infty)$. And it is trivial that $\lVert f\rVert_1 = 1$.
For (b), let us write generic vector as$$
v=(x,y,z,v_4,v_5,\ldots)=xe_1+ye_2+ze_3 +v_4e_4 +\cdots.$$ Note that for any $\{\alpha_j\}_{j\geq 4}$ with $\sup_{j\geq 4}|\alpha_j|\leq 1,$ $$
F(v) = x-y-z +\sum_{j=4}^\infty \alpha_j v_j$$ is a bounded extension of $f$ with $$
\lVert F\rVert_1 = 1=\lVert f\rVert_1.$$ This shows there are infinitely many non-expanding extensions for $p=1$. 
For the case $p\in (1,\infty)$, suppose $F:l_p(\mathbb{N})\to\mathbb{K}$ is a bounded extension of $f$ with $$\lVert F\rVert_p =3^{1 -\frac{1}{p}}.$$ Let $F(e_j) = \alpha_j$. We can see that for all $j\geq 4$, it holds that
$$\begin{eqnarray}
|F\left(xe_1+ye_2+ze_3 + w e_j\right)| = |x-y-z-\alpha_jw|&\leq& (|x|^p+|y|^p+|z|^p+|w|^p)^{\frac{1}{p}}\\&&\times(1+1+1+|\alpha_j|^{\frac{p}{p-1}})^{1 -\frac{1}{p}}\\&=&(3+|\alpha_j|^{\frac{p}{p-1}})^{1 -\frac{1}{p}}\lVert v\rVert_p
\end{eqnarray}$$ by Holder's inequality. Since equality can be attained, we have
$$
(3+|\alpha_j|^{\frac{p}{p-1}})^{1 -\frac{1}{p}}\leq \lVert F\rVert_p =3^{1 -\frac{1}{p}},\quad\forall j\geq 4.
$$ This shows $\alpha_j = 0$ for all $j\geq 4$, and it follows that
$$
F(v) = x-y-z
$$ for all $v\in l_p(\mathbb{N})$. Thus, there is a unique non-expanding extension of $f$ as desired.
