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Can we find the maximum value of a sequence by considering a differentiable function which has the same values as the sequence on integers. Is it true that if the maximum occurs at some value $a_m$ for some integer $m$, then the differentiable function would be would have an extremum in the interval $[m-1,m+1]$?

Formally, let $f:\mathbb{R} \to \mathbb{R}$ be an everywhere diffrentiable function, then if $\exists\ n \in \mathbb{Z} $ such that $ f(n)> g(m)\ \forall \ m \in \mathbb{Z}-\left\{n\right\}$, then $\exists\ c \in \mathbb{R}$ such that $f'(c)=0$ and $|c-n| \le1$

I tried to create a counter example by plotting different curves on the plane but I couldn't find one nor could I prove it. Is the above statement true/false?

Peter
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  • It's certainly true if the derivative is continuous. – Gerry Myerson May 09 '17 at 12:53
  • Can you point me to a source where it is proven and what if the derivative isn't continuous? – Peter May 09 '17 at 12:55
  • $f(n)>f(n-1)$ implies there is a point of positive derivative between $n-1$ and $n$. $f(n)>f(n+1)$ implies there is a point of negative derivative between $n$ and $n+1$. – lulu May 09 '17 at 13:00
  • @lulu what if the derivative is not continuous? – Peter May 09 '17 at 13:10
  • @Peter I was thinking only of the continuous case. – lulu May 09 '17 at 13:40
  • @Peter An everywhere differentiable function always has continuous derivative. If you relax the "everywhere differentiable" requirement on $f$, then the statement is clearly false, because $f$ might not be differentiable at the critical value. – Reese Johnston May 09 '17 at 13:43
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    @Reese, see https://math.stackexchange.com/questions/292275/discontinuous-derivative for discussions of everywhere differentiable functions with discontinuous derivatives. Also worth a look: https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be – Gerry Myerson May 10 '17 at 02:53
  • @GerryMyerson I stand corrected! – Reese Johnston May 10 '17 at 03:22

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Assume that $n\in\Bbb Z$ and $f(n)>f(m)$ for every $m\in\Bbb Z\setminus\{n\}$.

Now consider the restriction $$g(x)=f(x)\text{ for }x\in[n-1,n+1]$$

Then $g$ has a maximum (Weierstrass' theorem) and it is not $g(n-1)$ or $g(n+1)$, since $g(n)$ is greater than both. At this maximum, the derivative of $g$ and $f$ is zero.

Remark: this proves something slightly stronger. You can say (with your notation) that $|c-n|<1$.

ajotatxe
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