# A problem of functions

+ 0 like - 0 dislike
1974 views

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:

$$f(x)=e^x$$

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.

retagged Sep 4, 2020

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?

+ 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 (-80 points)
edited Sep 4, 2020

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 use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification