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Theorem 12.4.3 ( That rises from the book Analytic Inequalities and Their Applications in PDEs By Yuming Qin but no proof is given.)

Let $\Omega $ be a smooth open set in $\mathbb {R^{n}}$ and $f,g\in L^{p}(\Omega)$ with $1\leq p\leq +\infty.$ Then we have $$f+g\in L^{p}(\Omega)$$ and $$\left \| f+g \right \|_{ L^{p}(\Omega)}\leq\left \| f \right \|_{ L^{p}(\Omega)}+\left \| g \right \|_{ L^{p}(\Omega)}.\quad (12.4.9)$$

If $p=1$,then the equality in $(12.4.9)$ holds if and only if $\arg f(x)=\arg g(x)$ a.e.in $\Omega.$

If $p>1$,then the equality in $(12.4.9)$ holds if and only if there exist constants $C_{1}$ and $C_{2}$ which are not all zero such that $C_{1}f(x)=C_{2}g(x)$ a.e. in $ \Omega $ or there exists a non-negative measueable function $h$ such that $fh=g$ a.e. in the set $A=\{x \in\Omega|f(x)g(x)\ne0\}.$

But @Daniel Fischer♦ had proved that "For $1 < p < \infty$,$$\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$$

holds if and only if there are non-negative real constants $\alpha,\beta$, not both zero, such that $\alpha f(x) = \beta g(x)$ almost everywhere."see here

Which one is right ? Is the condition "non-negative" necessary for the equality holds when $p>1$? What does the symbol $\arg f(x)=\arg g(x)$ mean in Theorem 12.4.3 $p=1$?

Aleph Alpha
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  • Yes, non-negative is important. Compare with the $|a+b|=|a|+|b|$ for real or complex numbers $a$ and $b$. – Kavi Rama Murthy Feb 16 '21 at 10:03
  • @Kavi Rama Murthy: What does $\arg f(x)=\arg g(x)$ mean in Theorem 12.4.3 $p=1$? I am sorry to say that I have little knowledge about functional-analysis. Thanks for your reply! – Aleph Alpha Feb 16 '21 at 10:17
  • Do you know if there is a free access to the book by Yumming Qin? – Rem Feb 16 '21 at 10:19
  • They are considering complex valued functions. $arg z$ is the argument of the complex num ber $z$. – Kavi Rama Murthy Feb 16 '21 at 10:19
  • It would be very surprising to me, if the result you quoted were right. For example, for $p=2$ then $f(x)=e^{-x^2}, g(x)=e^{-(x-1)^2}$ satisfy the assumptions. Is it really the case that $|f+g|_2=2|f|_2$? – Giuseppe Negro Feb 16 '21 at 10:33
  • @GiuseppeNegro In your example $frac {f(x) {g(x)}$ is not a constant. – Kavi Rama Murthy Feb 16 '21 at 11:38
  • @AlephAlpha I am not the one who asked for a free copy of the book. It is illegal to download books from libgen. – Kavi Rama Murthy Feb 16 '21 at 11:39
  • @KaviRamaMurthy: but there is a function $h\ge 0$ such that $fh=g$. This is what I find surprising; that if this condition is satisfied, then $|f+g|=|f|+|g|.$ – Giuseppe Negro Feb 16 '21 at 12:55

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