8-3

Proof of Theorem 8.12, the Holder's inequality
Case 1: p=1, q = \infty, 
\begin{align}
\|a b\|_1 = \sum_k\left|a_k b_k\right| \leq\ (\sum_k\left|a_k \right|) \left|\sup _{k}\left|b_{k}\right|\right| = |a\|_1\|b\|_\infty
\end{align} 
Case 2: p = \infty, q = 1 ,
similar proof in case 1
Case 3: 1 < p, q < \infty
If ||a||_p=0 or ||b||_p=0, Holder's inequality is obvious holds.
For 0 < ||a||_p,||b||_p < \infty, take \|A\|_p = \|B\|_q = 1 
By Young's inequality,
\begin{align}
\|A B\|_1 = \sum_k\left|A_k B_k\right| & \leq \sum_k\left(\frac{\left|A_k\right|^p}{p}+\frac{\left|B_k\right|^q}{q}\right)=\frac{\|A\|_p^p}{p}+\frac{\|B\|_q^q}{q}=\frac{1}{p}+\frac{1}{q}=1 
\end{align} 
Let A=\frac{a}{\|a\|_p}, B=\frac{b}{\|b\|_p}
\begin{align}
\|a b\|_1 & =\sum_k\left|a_k b_k\right|=\|a\|_p\|b\|_q \sum_k\left|A_k B_k\right| \leq\|a\|_p\left\|b\right\|_{q}
\end{align}
Proof of Theorem 8.13, the Minkowski's inequality
Case 1: p=1 
\begin{align}
\|a+b\|_1 = a+b = \|a\|_1+\|b\|_1
\end{align} 
Case 2: p=\infty 
By triangle inequality,
\begin{align}
\|a+b\|_\infty 
&= \sup _{k}\left|a_{k} + b_{k}\right| \\
&\leq \sup _{k}\left|b_{k}\right| + \sup _{k}\left|b_{k}\right|\\
&= \|a\|_\infty + \|b\|_\infty
\end{align}
Case 3: 1 < p <\infty
\begin{align}
\|a+b\|_p^p & =\sum_k\left|a_k+b_k\right|^p \\
& \leq \sum_k\left|a_k+b_k\right|^{p-1}\left|a_k\right|+\sum_k\left|a_k+b_k\right|^{p-1}\left|b_k\right| \\
& \leq\left(\sum_k\left|a_k+b_k\right|^p\right)^{\frac{p-1}{p}}\left(\sum_k\left|a_k\right|^p\right)^{\frac{1}{p}}+\left(\sum_k\left|a_k+b_k\right|^p\right)^{\frac{p-1}{p}}\left(\sum_k\left|b_k\right|^p\right)^{\frac{1}{p}} \\
& =\|a+b\|_p^{p-1}\left(\|a\|_p+\|b\|_p\right)
\end{align}
Divide both sides by \|a+b\|_p^{p-1}, we get
\|a+b\|_{p} \leq\|a\|_{p}+\|b\|_{p}
A counter example for p < 1
Let a=\left\{a_{k}\right\}, b=\left\{b_{k}\right\},Suppose k_{0} \in \mathbb{Z}^{+}, k_{0}>1
\begin{array}{l}
a_{k}=\left\{\begin{array}{ll}
\frac{1}{2}, & \leqslant 1 \leq k \leq k_{0} \\
0, & k>k_{0}
\end{array} 
 b_{k}=\left\{\begin{array}{ll}
0, & 1 \leqslant k \leqslant k_{0} \text { or } k > 2 k_{0} \\
\frac{1}{2}, & k_{0}<k \leqslant 2 k_{0}
\end{array}\right.\right. \\
c=a+b=\left\{\frac{1}{2}\right\}_{k=1}^{2k_0} \\
\|a+b\|_{p}=\left(\sum_{k=1}^{2 k_{0}}\left(\frac{1}{2}\right)^{p}\right)^{\frac{1}{p}}=\left(2 k_{0}\right)^{\frac{1}{p}} \cdot \frac{1}{2} \\
\|a\|_{p}+\|b\|_{p}=\left(\sum_{k=1}^{k_0}\left(\frac{1}{2}\right)^{p}\right)^{\frac{1}{p}}+\left(\sum_{k=k_0+1}^{2k_0}\left(\frac{1}{2}\right)^{p}\right)^{\frac{1}{p}}=2\left(\sum_{k=1}^{k_{0}}\left(\frac{1}{2}\right)^{p}\right)^{\frac{1}{p}}=2 \cdot k_{0}^{\frac{1}{p}} \cdot \frac{1}{2} \\
\end{array}
Since p<1, 2^{\frac{1}{p}}>2 \Rightarrow\left(2 k_{0}\right)^{\frac{1}{p}}>2 k_{0}^{\frac{1}{p}}