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please I tried to find counterexamples to see that $l^p$ is not norm with $p<1$ in the triangle inequality but I have problems with convergence when I choose some successions. Thanks.

ajotatxe
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mathreda
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    Let $p=1/2$ and $a,b>0$ Then $(\sqrt{a} + \sqrt{b})^2=a+b+2\sqrt{ab}> a+b$.

    Let $x \in l^p$ be defined as $x_1=a$ and $x_i=0$ for $i\neq 1$, and let $y \in l^p$ be defined as $y_2=a$ and $y_i=0$ for $i\neq 2$. Then we have $$|x+y|_p=(\sqrt{a} + \sqrt{b})^2 =a+b+2\sqrt{ab}> a+b = |x|_p+ |y|_p$$

     – Ramiro Sep 08 '16 at 00:39

2 Answers2

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Let $p<1$. Then

$$\|(1,0,0,\ldots,0\ldots)\|_p=1$$

$$\|(0,1,0,\ldots,0\ldots)\|_p=1$$

$$\|(1,1,0,\ldots,0\ldots)\|_p=2^{1/p}>2$$

ajotatxe
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3

The triangle inequality is violated for $v_1=(1,0,0,...)$ $v_2=(0,1,0,0,...)$.

Paul
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