Optimal coefficient in Cauchy-Schwartz inequality?

Question

This may be trivially wrong, but I can't see a counterargument. Let $f,g:\mathbb{R}\to\mathbb{R}$ be measurable complex-valued functions such that $f/g$ is defined. Does there exist a coefficient $\alpha(f/g)$ depending only on their quotient as a measurable function such that $$ |(f,g)|\le\alpha(f/g)\|f\|\|g\|. $$ $\alpha(h)=1$ if $h=\mathrm{const}$ (different from 0), so maybe $\alpha$, if it exists, has to do with the variance. Thank you.

Answer 1

If $h$ is not equal to zero almost everywhere, then $\alpha(h)=1$.

First let us restrict ourselves to $f,g \in L^2(\mathbb{R})$. Otherwise we run into troubles with multiplying zero and infinity. I am assuming that we are talking about the Lebesgue measure with the usual Lebesgue sigma-algebra.

Case 1: h equal to zero almost everywhere. Let's take care of the trivial case. If $h$ is almost everywhere equal to zero, then we have $f=h$ a.e. and $g$ some a.e. positive $L^2(\mathbb{R})$ function. Then we have $$ \vert (f, g) \vert = 0 $$ As also $\Vert f \Vert=0$, we can pick $\alpha(h)$ whatever we like (even negative constants).

Case 2: if h is not equal to zero almost everywhere. Let $h: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function such that $h$ is not almost everywhere equal to zero. First we note that there always exist admissible $L^2(\mathbb{R})$ functions such that $h=f/g$, for example if $H\in L^2(\mathbb{R})$ positive, then $$ f= \left(\frac{1_{\{\vert h \vert \geq 1\}}}{h} + 1_{\{\vert h\vert <1\}} h \right) H, \qquad g= \left(\frac{1_{\{\vert h \vert \geq 1\}}}{h^2} + 1_{\{\vert h\vert <1\}} \right) H $$ do the job. We have $$ \alpha(h) = \sup_{f,g\in L^2(\mathbb{R}): h=f/g} \frac{\vert (f,g) \vert}{\Vert f \Vert \cdot \Vert g\Vert}.$$ Note that $\Vert f \Vert, \Vert g \Vert \neq 0$ as $h$ is not equal zero almost everywhere. Furthermore, by Cauchy-Schwarz we know that $\alpha(h) \leq 1$. Hence, we just need to show that $\alpha(h)\geq 1$.

Handwaving outline: Our strategy is the following. We can replace $f,g$ by $1_A f, 1_A g$. Then we will pick $A$ to be roughly a level set of $h$. In that case we have that $1_Af, 1_A g$ are almost scalar multiples of each other. Then we get that the quotient is roughly one.

Actual proof: In general we have for $C>0$ and $A\subseteq \mathbb{R}$ measurable set not being a null set, that if $f,g$ are admissible, then so are $1_A f + \frac{1}{C} 1_{A^c} f $, $1_A g + \frac{1}{C} 1_{A^c} g $.

Hence, if we let $C$ go to infinity, we get for every $A\subseteq \mathbb{R}$ measurable set such that $\{ h \neq 0 \}\cap A$ is not a null set (and thus $\Vert 1_A f \Vert, \Vert 1_A g\Vert \neq 0$) $$ \alpha(h) \geq \sup_{f,g\in L^2(\mathbb{R}): h=f/g} \frac{\vert (1_A f,1_A g) \vert}{\Vert 1_A f \Vert \cdot \Vert 1_A g\Vert}. $$ Now we want to make $A$ almost a level set of $h$ such that $1_A f$ and $1_A g$ are almost scalar multiples of each other.

First we note that $\alpha(h) = \alpha(-h)$, hence, we wlog of generality we may assume that $\{ h>0\}$ is not a null set. Thus, we can pick $R>0$ such that $\{h>0 \}\cap [-R,R]$ is not a null set. For every $\varepsilon >0$ there exists $\xi(\varepsilon)>0$ such that $$A_\varepsilon := \{h>0 \} \cap h^{-1}([\xi(\varepsilon) -\varepsilon, \xi(\varepsilon) + \varepsilon]) \cap [-R, R]$$ is not a null set. By subdividing the interval $[\xi(\varepsilon) -\varepsilon, \xi(\varepsilon) + \varepsilon]$ we find a sequence $(\xi_{n_l})_{l \geq 1}$ ($n_l < n_{l+1}$) such that $\xi_{n_l}\geq 1/\sqrt{n_l}$ and such that $$ B_{n_l} := h^{-1}([\xi_{n_l} -1/n_l, \xi_{n_l} + 1/n_l]) \cap [-R, R] $$ is not a null set. Now, if we pick $A=B_{n_l}$ and $f=h$, $g=1$ on $B_{n_1}$ (this we are allowed to do as $B_{n_1}$ has finite measure as it is a subset of $[-R,R]$ and $h$ is bounded on that set) we get $$ \alpha(h) \geq \frac{\vert (1_{B_{n_l}} h,1_{B_{n_l}} 1) \vert}{\Vert 1_{B_{n_l}} h \Vert \cdot \Vert 1_{B_{n_l}} g\Vert} \geq \frac{(\xi_{n_l} - 1/n_l) \Vert 1_{B_{n_l}} \Vert^2}{(\xi_{n_l} + 1/n_l) \Vert 1_{B_{n_l}} \Vert^2} = \frac{(\xi_{n_l} - 1/n_l)}{(\xi_{n_l} + 1/n_l)} \geq \frac{(\xi_{n_l} - 1/n_l)}{\xi_{n_l}} \geq \frac{(1-1/\sqrt{n_l}) \xi_{n_l}}{\xi_{n_l}}= (1-1/\sqrt{n_l}).$$ Let $l$ go to infinity, then we obtain $$ \alpha(h) \geq 1.$$

Added: It was pointed out in the comments that it is not obvious that we can choose $\xi_{n_l}\geq 1/\sqrt{n_l}$. Let me elaborate on that. We know that there exists some $L>0$ such that $$ A := \{h>0 \} \cap h^{-1}((0,L]) \cap [-R, R] $$ has positive measure. Writing $$ A = \bigcup_{n\geq 1} \left( \{h>0 \} \cap h^{-1}((L/2^{n},L/2^{n-1}]) \cap [-R, R] \right) $$ we get that there exists $\xi >0$ such that $$ \{h>0 \} \cap h^{-1}([\xi,L]) \cap [-R, R] $$ has positive measure. Now pick $n_1$ such that $\xi \geq 1/\sqrt{n_1}$. We have for $N\geq n_1$ $$ \{h>0 \} \cap h^{-1}([\xi,L]) \cap [-R, R] = \bigcup_{j=0}^{N^2-1}h^{-1}([\xi + j\frac{L-\xi}{N^2}, \xi + (j+1) \frac{L-\xi}{N^2} ])\cap [-R, R] $$ Then one of those sets will have positive measure and the length of the interval is $\frac{L-\xi}{N^2}$ and the midpoints of the intervals are greater or equal to $\xi$. Hence, we in particularly pick $\xi_n$ in the way I wrote above (in fact I can make the length of the intervals to decay as fast as I want and the inequality of the midpoints versus the length could be taken as good as you would like).