Let $0< p< 1$ and let $\|f\| _{p}$ be defined as $(\int_{0}^{1} |f|^p)^{1/p}.$ Prove or disprove: $\|\cdot\|_{p}$ is a norm on $L^{p}([0,1])$
My trial:
It is a norm, but I am unable to prove the triangle inequality? Could anyone help me in this?
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This is not a norm. Take $f=I_A, g=I_B$ where $A$ and $B$ are disjoint sets of positive measures $a$ and $b$. Then $\|f+g\|_p =(a+b)^{1/p}$, $\|f\|_P =a^{1/p}$ and $\|g\|_P =b^{1/p}$ But $(a+b)^{1/p} \leq a^{1/p} +b^{1/p}$ is false. In fact $(a+b)^{1/p} > a^{1/p} +b^{1/p}$.
Kavi Rama Murthy
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$I_{A}$ is the identity function on $A$ correct? – Emptymind Nov 17 '19 at 01:22
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1It is 1 on A and 0 elsewhere. – Kavi Rama Murthy Nov 17 '19 at 02:03
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could you please tell me the details of calculating this $p-$norm please? – Intuition Dec 03 '19 at 04:12
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specially the $p-$norm of $f + g$ – Intuition Dec 03 '19 at 04:18
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why are you taking $A$ and $B$ disjoint? – Intuition Dec 03 '19 at 04:19
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Also our $E$ is the interval [0,1] not $A$ or $B$, so why did you take $a, b$? – Intuition Dec 03 '19 at 04:22
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@Secretly $a$ and $b$ are not end points of the interval. They just any numbers between $0$ and $1$. – Kavi Rama Murthy Dec 03 '19 at 05:29
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and why $A$ and $B$ are disjoint? – Intuition Dec 03 '19 at 10:05
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@Secretly I cannot say $|f+g|_p=(a+b)^{1/p}$ if they are not disjoint. – Kavi Rama Murthy Dec 03 '19 at 10:07
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So could you please tell me on which theorem you based your calculation for this step? – Intuition Dec 03 '19 at 10:09
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1$\int |f+g|^{p}=\int_A |f+g|^{p}+\int_B |f+g|^{p}+\int_{(A\cup B )^{c}} |f+g|^{p}=a+b$ – Kavi Rama Murthy Dec 03 '19 at 10:13