Suppose $f_n,g_n:\mathbb{R} \to \mathbb{R}$ are smooth functions and $f_n - g_n$ has a maximum point at $x_n$ ($\forall n$).

Then we have $$f'_n(x_n) = g_n'(x_n)$$ $$f_n''(x_n) \le g_n''(x_n).$$

Now let $\epsilon>0$ and $\delta \in \mathbb{R}$ be small numbers.

If we choose $\delta$ well enough (compared to $\epsilon$), is it true that

$$\epsilon f_n''(x_n) + \delta f_n'''(x_n) \le \epsilon g_n''(x_n) + \delta g_n'''(x_n) \ ?$$ (where the $n$-th terms of the sequences depend on $\epsilon$ and $\delta$) for $\epsilon$ and $\delta$ sufficiently small?

Otherwise, what other reasonable assumptions do we need on $\{g_n\}$ and $\{f_n\}$ to make the result hold true?


It is clear that $\delta=0$ works. If you want $\delta>0$ then it may be not possible to do. Let $f_n(x)=-x^2+n\,x^3$ and $g_n(x)=0$. Then $f_n-g_n=f_n$ has a maximum at $x_n=0$, and for $\epsilon,\delta>0$ we have $$ \epsilon\,f_n''(0)+\delta\,f_n'''(0)=-2\,\epsilon+6\,n\,\delta. $$ If this has to be $\le0$ for all $n$, then $\delta\le\epsilon/(3\,n)$ for all $n$, which is not possible.

From this example, we see that a sufficient condition is $$ \inf_{n}\frac{f_n''(x_n)-g_n''(x_n)}{f_n'''(x_n)-g_n'''(x_n)}>0. $$

  • $\begingroup$ Thank you. When I asked the question I had in mind a sufficient condition of "relative smallness" of $\epsilon$ and $\delta$, that is something like, say, $\delta = O(\epsilon^2)$ as they go to $0$. Is it possible to obtain the result under such condition? $\endgroup$ – user411609 Jun 27 '17 at 12:16
  • $\begingroup$ In the above example, given $\epsilon>0$, there is no $\delta>0$ satisfying the condition. The answer seems to be no. $\endgroup$ – Julián Aguirre Jun 28 '17 at 11:02

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