User:J D Bowen/Math711

1)     For $||x||_1$, observe that since this is a finite sum of absolute values, and since addition is componentwise and scalar multiplication distributes across components, and since absolute value satisfies all the conditions of a norm, $||x||_1$ is a norm.

Observe that for $||x||_p$, we have

$$||ax||_p = \left({ \sum_{j=1}^n |ax_j|^p }\right)^{1/p} = \left({ |a|^p \sum_{j=1} |x_j|^p }\right)^{1/p}=|a| ||x||_p$$

and

$$||x+y||_p = \left({ \sum_{j=1}^n |x_j+y_j|^p }\right)^{1/p} \leq \left({ \sum_{j=1}^n (|x_j|+|y_j|)^p }\right)^{1/p} \leq \left({ \sum_{j=1} |x_j|^p }\right)^{1/p}+\left({ \sum_{j=1} |y_j|^p }\right)^{1/p} = ||x||_p+||y||_p$$

Finally, we have

$$||x||_p = 0$$

$$\implies \left({ \sum_{j=1} |x_j|^p }\right)^{1/p} = 0$$

$$\implies \sum_{j=1} |x_j|^p =0$$

Since $\forall j, |x_j|\geq 0$, this final line line implies $|x_j|^p=0$ for all $j$, and hence $x_j=0, \implies x=0$.

For the final norm, we have

$$||ax||_\infty = \text{max}_{1\leq j \leq n} |ax_j|  = \text{max}_{1\leq j \leq n}  |a||x_j| =|a| \text{max}_{1\leq j \leq n}  |x_j|  =|a| ||x||_\infty $$

We also note $$||x+y||_\infty = \text{max}_{1\leq j \leq n} |x_j+y_j| \leq \text{max}_{1\leq j \leq n} |x_j|+|y_j| \leq \text{max}_{1\leq j \leq n} |x_j|+\text{max}_{1\leq j \leq n} |y_j|= ||x||_\infty+||y||_\infty$$

Finally, observe that $||x||_\infty =0\implies \text{max}|x_j|=0 \implies \forall j, x_j=0 \implies x=0$.

2) The drawing in the text shows the relation of $||x||$ being a circle, $||x||_\infty$ being a square, and the unit circles for $||x||_p$, which we call $C_p$, being curves between the two with $m>n, x\in C_m, y \in C_n \implies ||x|| \geq ||y||$.

Suppose

$$\left({ \sum_{j=1} |x_j|^m }\right)^{1/m}=\left({ \sum_{j=1} |y_j|^n }\right)^{1/n}=1$$

Then

$$ \sum_{j=1} |x_j|^m = \sum_{j=1}^n |y_j|^n =1$$,

since $1^m=1^n=1$. But since all coordinates must be $|x_j|\leq 1$ to solve this equation, and since in this range $m>n \implies |x_j|^m \leq |x_j|^n$, we have

$$\sum_{j=1} |y_j|^n \leq \sum_{j=1} |x_j|^n$$

Taking the nth root of both sides we get $||y||\leq ||x||$ as desired, under any metric of the form $||*||_p$.

3) Define $\phi(x)=(\sqrt{|x_1|}+\sqrt{|x_2|})^2=|x_1|+|x_2|+2|x_1x_2|$. Observe that for $x=(1,1)$, we have $\phi(x)=4$, but $\phi(x+x)=12$, so this fails to satisfy the triangle inequality.