Prove or disprove $d(f,g):= \lVert f-g\rVert_p^p$ defines a metric on $L^p(\mu)$ for $0

Question

Prove or disprove: $d(f,g):= \lVert f-g\rVert_p^p$ defines a metric on $L^p(\mu)$.

My attempt:

Use the fact that for $x,y\geq0$ and $0<p<1$, $$(x+y)^p\leq x^p+y^p$$ implies $$\int\vert{f+g}\vert^pd\mu\leq\int\vert f \vert^pd\mu+\int\vert g \vert^pd\mu$$ from which we have:$$d(f,-g)\leq d(f,0)+d(0,-g)$$ I can only prove the triangle inequality for $0,f,g$. To define a metric on $L^p(\mu)$, we need the triangle inequality for any $h,f,g \in L^p(\mu)$.

Thanks in advance.

@JesseMadnick Hi, I am only interested in the case $0<p<1$, I know this is false for $p\geq1$ — Flashhh, Jun 29 '21 at 08:07
Here I another related question that also shows that $|f-g|^p_P$ in $L_p$ $(0<p<1)$ defines a complete metric. — Mittens, Jun 29 '21 at 08:15

score 3 · Accepted Answer · answered Jun 29 '21 at 07:59

3

$f-g =(f-h)+(h-g)$ so $|f-g|\leq |f-h|+|h-g|$ and $|f-g|^{p} \leq (|f-h|+|h-g|)^{p}$. Take $x=|f-h|, y=|h-g|$ in the inequality $(x+y)^{p} \leq x^{p}+y^{p}$.

answered Jun 29 '21 at 07:59

Kavi Rama Murthy

311,013

The last inequality is clearly wrong. Take $p=2$ and we have $(x+y)^2 = x^2+y^2 +2xy > x^2+y^2$ if $x\cdot y >0$. – Kernel Jun 29 '21 at 08:09
@Kernel Hi, this inequality is just for the case $0<p<1$ – Flashhh Jun 29 '21 at 08:09
Oh, sorry, I didn't have the second comment loaded yet! – Kernel Jun 29 '21 at 08:11

Prove or disprove $d(f,g):= \lVert f-g\rVert_p^p$ defines a metric on $L^p(\mu)$ for $0

1 Answers1