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  A problem of functions

+ 0 like - 0 dislike

I consider a smooth function $f$ over the real numbers, the derivatives are $f^{(k)}$. I suppose that:

$$0< f^{(n+1)}(x) \leq f^{(n)}(x)$$

for all $n \in {\bf N}$ and all $x \in {\bf R}$. Then I suppose that for $x \cong -\infty$, $f (x)\sim e^x$, ($f$ is equivalent to $e^x$ in $-\infty$); have we in these conditions:


This problem may have a meaning in physics because if $t$ is the time, $x=\ln(t)$, we have near the Big Bang $t=0$, $x\cong -\infty$, so that $f$ is in fact the time.

asked Aug 31, 2020 in Mathematics by Antoine Balan (-80 points) [ no revision ]
retagged Sep 4, 2020 by Antoine Balan

Most likely not. What is your definition of equivalence?

I take $f \sim_x g$ if and only if $\lim_x f/g=1$. It would help if you had a concrete counter-example. It is not a question of probability or philosophy. Moreover if $f$ is a sum of exponential functions, it is true.

I first thought that one can construct counterexample of the form $f(x)=e^xf(x^2)/g(x^2)$ with low degree polynomials $f,g$ of the same degree and the same highest coefficient. Your inequalities specify relations between the coefficients. Your conjecture is that these have the zero solution only. But the global positivity of all derivatives is a very strong condition, and polynomials will probably cause too much oscillations. So I retract my pessimistic first assessment.

What is the physical relevance of your inequalities?

Indeed, what it is all about?

1 Answer

+ 2 like - 0 dislike

The problem can be solved by the Bernstein theorem about total monotony of functions. Because $f^{(n)}(x)\geq 0$, the Bernstein theorem implies that $f(x)=\int_0^{+\infty} e^{xt} dg$ for $x<0$, with $g$ an increasing function. As $f^{(n+1)}(x) \leq f^{(n)}(x)$, we have $\int_0^{+\infty} t^n (t-1) dg \leq 0$, so that $g(t)=K$, with $K$ fixed, for $t>1$. Now, $f(x)\sim e^x$ for $x\cong -\infty$ implies that $g(t)=K'$ for $t<1$.

answered Sep 2, 2020 by Antoine Balan (-80 points) [ revision history ]
edited Sep 4, 2020 by Antoine Balan

This is not an answer but only a hint (and it includes a typo). How is it solved?

The Bernstein theorem implies that $f(x)=\int_0^{+\infty} e^{xt} dg$ for $x<0$ and $g$ an increasing function (as $f^{(n)}(x)\geq 0$). Then $f^{(n+1)}(x)\leq f^{(n)}(x)$  implies that $\int_0^{+\infty} e^{xt}t^n (t-1)dg \leq 0$ so that $g(t)=K$, with $K$ fixed, for $t>1$. As $f(x)\sim e^x$, $g(t)=K'$ for $t<1$.

Please put the argument into the answer.

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