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Consider the inequality $$\left(\left(\int{f}\right)^{p}+\left(\int{g}\right)^{p}\right)^{1/p} \le \int{(f^p+g^p)^{1/p}}$$ where $f$ and $g$ are nonnegative measurable functions and $p \ge 1$ .

I believe that I can show it's true by MCT and simple functions, but what is its equality case?

My guess is that f and g are multiples up to a null set, but I'm not sure how to prove this.

davik
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  • try if either is zero – mathreadler Jun 14 '17 at 18:34
  • I would look into Jensen's inequality. With some effort I think you can use it to conclude not only the inequality but also that equality only holds when $f$ and $g$ are constant, or $fg = 0$ almost everywhere. – Contravariant Jun 14 '17 at 20:03
  • In the case $p=1$ we always have equality. – md2perpe Jun 14 '17 at 21:09
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    If we define $h : { 0, 1 } \times X \to \mathbb R_+$ through $h(0, x) = f(x)$ and $h(1, x) = g(x)$ and on ${ 0, 1 }$ take the counting measure, then the inequality can be written $\left| | f(k, x) |{L^1_x} \right|{L^p_k} \leq \left| | f(k, x) |{L^p_k} \right|{L^1_x}$. Here $k$ and $x$ are used inside the norm just to mark in what position the nested norms are taken. I wonder if it's generally the case that $\left| |f(x,y)|{L^p_x} \right|{L^q_y} \leq \left| |f(x,y)|{L^q_y} \right|{L^p_x}$ when $1 \leq p < q$. – md2perpe Jun 14 '17 at 21:21
  • Yes my guess is only for $p>1$ – davik Jun 15 '17 at 15:30

1 Answers1

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Take $f=F^{1/p}$ and $g=G^{1/p}$. Then you get $$\left(\int F^{1/p}\right)^p+\left(\int G^{1/p}\right)^p\le \left(\int (F+G)^{1/p}\right)^p.$$ Now take $q=1/p$. Then $$\left(\int F^q\right)^{1/q}+\left(\int G^q\right)^{1/q}\le \left(\int (F+G)^q\right)^{1/q}.$$ Hence, what you have is the reverse Minkowski's inequality for $0<q<1$. To prove it you first prove that $$ab\ge \frac{a^q}q-\frac{b^{-r}}{r}$$ for $a\ge 0$ and $b>0$ where $\frac1q-\frac1r=1$. If $a>0$ and $b>0$ then you have equality iff $a^q=b^{-r}$. Using this you prove the reverse Holder inequality $$\int FG\ge \left(\int F^q\right)^{1/q}\left(\int G^{-r}\right)^{-1/r}$$ where $G(x)>0$ for every $x$. Again you have equality iff $(G(x))^{-r}=c(F(x))^q$ for a.e. $x$. Then you prove Minkoski's inequality in the usual way \begin{align}\int (F+G)^q&=\int_{\{F+G>0\}} (F+G)^q=\int_{\{F+G>0\}} (F+G) (F+G)^{q-1}\\&=\int_{\{F+G>0\}} F(F+G)^{q-1}+\int_{\{F+G>0\}} G (F+G)^{q-1}\\ &\ge \left(\int F^q\right)^{1/q}\left(\int (F+G)^{-r(q-1)}\right)^{-1/r}+\left(\int G^q\right)^{1/q}\left(\int (F+G)^{-r(q-1)}\right)^{-1/r}\\&=\left[\left(\int F^q\right)^{1/q}+\left(\int G^q\right)^{1/q}\right]\left(\int (F+G)^{q}\right)^{-1/r}.\end{align} As in the standard case you equality p>1 you get $F^q=c(F+G)^q$ a.e. and $G^q=d(F+G)^q$ a.e. so $F=aG$ a.e. I skipped a lot of details.....

Gio67
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