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Let $X=K^3$. For $x=(x(1),x(2),x(3))\in X$, let $||x||=[(|x(1)|^2+|x(2)|^2)^\frac{3}{2}+|x(3)|^3]^\frac{1}{3}$. Then $||.||$ is a norm on $K^3$

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2 Answers 2

Hint: note that $\|x\| = \|(r, x(3))\|_3$ where $r = \|(x(1),x(2))\|_2$.

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I assume that $K$ is a subfield of $\Bbb C$, here. Observe that we're dealing with a composition of various norms, in particular: the modulus function on $\Bbb C$, given by $|a+bi|=\sqrt{a^2+b^2}$ for all real $a,b$; the euclidean norm on $\Bbb C^2$, given by $\lVert(z_1,z_2)\rVert_2=\sqrt{|z_1|^2+|z_2|^2}$ for $z_1,z_2\in\Bbb C$; and the $3$-norm on $\Bbb C^2$, given by $\lVert(w_1,w_2)\rVert_3=\sqrt[3]{|w_1|^3+|w_2|^3}$ for $w_1,w_2\in\Bbb C$. Our given norm-hopeful is then $$\lVert(x_1,x_2,x_3)\rVert=\bigl\lVert\bigl(\lVert(x_1,x_2)\rVert_2,x_3\bigr)\bigr\rVert_3.$$

We can use norm properties of $|\cdot|$, $\lVert\cdot\rVert_2$, and $\lVert\cdot\rVert_3$ to show scaling and that $\lVert x\rVert=0$ implies $x=0$. Using norm properties of $\lVert\cdot\rVert_2$ and $\lVert\cdot\rVert_3$, along with the fact that $\lVert(w_1,w_2)\rVert_3\geq \lVert(w_1',w_2)\rVert_3$ whenever $|w_1|\geq|w_1'|$ (which you should be able to prove), we can also prove triangle inequality.

Explicitly, we have $$\begin{align}\bigl\lVert(z_1,z_2,z_3)+(w_1,w_2,w_3)\bigr\rVert &= \bigl\lVert(z_1+w_1,z_2+w_2,z_3+w_3)\bigr\rVert\\ &= \bigl\lVert\bigl(\lVert(z_1+w_1,z_2+w_2)\rVert_2,z_3+w_3\bigr)\bigr\rVert_3\\ &= \bigl\lVert\bigl(\lVert(z_1,z_2)+(w_1,w_2)\rVert_2,z_3+w_3\bigr)\bigr\rVert_3\\ &\le \bigl\lVert\bigl(\lVert(z_1,z_2)\rVert_2+\lVert(w_1,w_2)\rVert_2,z_3+w_3\bigr)\bigr\rVert_3\\ &=\bigl\lVert\bigl(\lVert(z_1,z_2)\rVert_2,z_3\bigr)+\bigl(\lVert(w_1,w_2)\rVert_2,w_3\bigr)\bigr\rVert_3\\ &\le\bigl\lVert\bigl(\lVert(z_1,z_2)\rVert_2,z_3\bigr)\bigr\rVert_3+\bigl\lVert\bigl(\lVert(w_1,w_2)\rVert_2,w_3\bigr)\bigr\rVert_3\\ &= \lVert(z_1,z_2,z_3)\rVert+\lVert(w_1,w_2,w_3)\rVert.\end{align}$$

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I think the triangle is difficult rest of the conditions can easily be proven!! – user44948 Dec 1 '12 at 16:55

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