Proving a equality using Cauchy-Schwarz inequality?

Asked May 19 '20 at 11:40

Active May 19 '20 at 12:21

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Could someone tell me if the following property is true?

$f,g:\mathbb{R}^N\to\mathbb{\mathbb{R}^N}$ satisfies that $\displaystyle\int_{\mathbb{R}^N}f\cdot g=\displaystyle\int_{\mathbb{R}^N}|f||g|$ if, and only if, $\exists\alpha:\mathbb{R}^N\to \mathbb{R}^+$ such that $f(x)=\alpha(x)g(x)$ a.e.

The right-to-left implication is evident

For left-to-right implication, I have tried to use, in some way, the condition for equality in Cauchy-Schwarz inequality. But I have not completed the proof.

If this property is not true, does anyone know a condition for which an equivalence as above is true?

edited May 19 '20 at 12:21

asked May 19 '20 at 11:40

prosep

Essentially the same argument – Sahiba Arora May 19 '20 at 11:44
Study the proof of Cauchy-Schwarz. It uses the fact that the integral over some non-negative function is $\ge 0$. What can you conclude if that integral is zero? – Martin R May 19 '20 at 11:48
Sorry, but I don't understand wath you are suggesting. Could you be a bit more explicit? – prosep May 19 '20 at 12:06
Is there a complex conjugate missing somewhere or something? Because taking $f$ and $g$ equal, where they have value $i$ on some measurable set and value $0$ elsewhere, satisfies the right part but not the left. – aschepler May 19 '20 at 12:18
Oh, sorry the functions take values in $\mathbb{R}^N$ not in $\mathbb{C}$. – prosep May 19 '20 at 12:21
I see. Then yes, since C-S gives that $f(x) \cdot g(x) \leq |f(x)||g(x)|$ for each $x$, look at the integral $\int ( |f||g| - f \cdot g )$. – aschepler May 19 '20 at 12:25
In fact, equality in the Cauchy-Schwarz inequality implies the existence of a scalar $\alpha \in \mathbb{R}_+$ such that $f = \alpha g$. – Michh May 19 '20 at 14:04

Proving a equality using Cauchy-Schwarz inequality?

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