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In trying to solve some quantum mechanical problem, I came across a proposition which can be formulated like the following:

$$\left|\int F f + G g \; d \tau' \right| \ge \left| \int F g - G f\; d \tau' \right|$$ for

$$ \left|\int F f d \tau' \right| \gg \left| \int G g \; d \tau' \right|, $$

with everywhere real, differentiable and integrable functions $F, f, G, g$. Is the statement possibly correct and in case how to show that?

I have something like Cauchy-Schwarz in mind, but I am quite inexperienced with inequalities and hope that some here might immediately see its wrong or not.

A simple test for numbers instead of functions and integrals would not contradict the inequality: $Ff=1, Gg=-\epsilon$ yields $|1-\epsilon| \ge |\epsilon\frac{f}{g}|$.

Remark 1: Some details about which I am not sure if they are of importance in the first instance are that all these functions lets call them $\psi$ depend on the set $\tau$ of variables: $\psi(\tau)$, while integration in the inequalities takes place only in the subset $\tau'\subset\tau$. The inequalities then are meant to hold point-wise for the remaining "parameter". Also $\int F d\tau = \int G d\tau = 1$, while $g$ and $f$ are the result of acting some differential operator on $F$ and $G$.)

Remark 2: To me it appears, that in this formulation its a mathematical rather then physical problem.

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  • $\begingroup$ I searching for counterexamples I have tested $F=-A,\,f=(x+1)(x-1),\,G=x^2,\,g=x^n$ for $\int_{-1}^{1}dx$. It gives $|\frac{2A}{3}+\frac{2}{2n+1}|>|\frac{2A}{n+1}-\frac{4}{15}|$, which looks good, I'd say, (with sufficiently large $A$ and $n$). $\endgroup$ – Rudi_Birnbaum Dec 10 '17 at 9:48
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The proposed inequality does not hold in this general form. A counterexample is $F(x) = -A,\; f=(x+1)(x-1),\; g=B,\; G=-x^{2n+1}$, for large $A,B\in\mathbb{R_+}$ and $n\in\mathbb{N_+}$, and integration in $[-1;1]$this yields: $$ \left| \int^{1}_{-1} -A(x+1)(x-1)\; -\; Bx^{2n+1} dx \right| \ge \left| \int^{1}_{-1} -AB\; +\; x^{2n+1}(x+1)(x-1) dx \right| $$
and $$ \left| \frac{2}{3}A \right| \ge \left| AB \right|. $$ Which apparently is wrong, while the condition $\left| \int_{-1}^{1} Ff\, dx \right| = \left|\frac{4A}{3} \right| \gg 0 = \left| \int_{-1}^{1} Gg\, dx \right|$ holds.

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