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$$f(x)=\lim_{m\to\infty}\frac{\sum_{n=-m}^m(x-n)^{1/3}}{\sum_{n=-m}^m(1-n)^{1/3}}\stackrel{?}{=}x$$

I had a 'proof' but I made the simple mistake of assuming 2 limits could be swapped, so I only really proved that $f(x)=x$ for $x\in[-1,0,1]$, which is pretty trivial. Here's the 'proof' that I had. This was a first draft so it may be confusing to read.

The $(\text{stuff})^{1/3}$ in the numerator and denominator is really just $g(\text{stuff})$. Where $g$ just needs to be any odd function and relatively smooth and have a few other properties that I don't remember.

vitamin d
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Jacob Claassen
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    Suggestion: the bottom is simply $m^{1/3}+(m+1)^{1/3}$, while the top can be bounded above and below by integrals (since the summand is a decreasing function of $n$ for any fixed $x$). – Greg Martin Jul 23 '21 at 02:13
  • @GregMartin, it is, but I like to keep it as the sum since it shows that it's only there as a term to keep the sum with "$x$" in it from diverging at x=1. Also I tried defining it with integrals but it was too complex to solve, especially in a general case with the arbitrary $g$ function mentioned. – Jacob Claassen Jul 25 '21 at 02:51
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    Could you add at least what you have tried even after “forgetting some properties”? I do have a generalizable proof but I refrain from posting it because the question, as it stands currently, is somewhat below the borderline suggested in Enforcement of Quality Standards. – Ѕᴀᴀᴅ Jul 25 '21 at 11:06
  • @Saad, I found the incorrect proof. https://pdfhost.io/v/MgJqCP5DR_Periodic_Sumstex.pdf – Jacob Claassen Jul 25 '21 at 17:04

1 Answers1

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We prove the following claim:

Claim. Suppose $g : \mathbb{R} \to \mathbb{R}$ satisfies the following condition:

  1. $g$ is an odd function.

  2. There exists $r \geq 0$ such that $g$ is either convex or concave on $[r, \infty)$.

  3. $\displaystyle \lim_{x\to\infty} |g(x)| = \infty$.

  4. For any $a \in \mathbb{R}$, $\displaystyle \lim_{x\to\infty} g(x+a)/g(x) = 1 $.

Then for any $x \in \mathbb{R}$,

$$ f(x) = \lim_{m\to\infty} \frac{\sum_{n=-m}^{m} g(x-n)}{\sum_{n=-m}^{m} g(1-n)} = x. $$

For example, $g(x) = \operatorname{sign}(x)|x|^p$ works for any $p > 0$, and this includes $g(x) = x^{1/3}$ as a special case.

Proof. By replacing $g$ by $-g$ if necessarily, we may assume that $g$ is convex on some $[r, \infty)$. also, it suffices to consider the case $x > 0$. Write

$$ G_m(x) = \sum_{n=-m}^{m} g(x-n). $$

Fix a positive integer $N$ such that $N > r + x + 1$. Then for $m \geq N$,

$$ G_m(x) = G_{N-1}(x) + \sum_{n=N}^{m} (g(n+x) - g(n-x)). $$

If $n \geq N$, then $g$ is convex on $[n-x-1, \infty) \subseteq [r, \infty)$, and so,

$$ g(n-x) - g(n-x-1) \leq \frac{g(n+x) - g(n-x)}{2x} \leq g(n+x+1) - g(n+x). $$

So it follows that

\begin{align*} &G_{N-1}(x) + 2x(g(m-x) - g(N-x-1)) \\ &\quad\leq G_m(x) \\ &\qquad \leq G_{N-1}(x) + 2x(g(m+x+1) - g(N+x)) \end{align*}

Dividing both sides by $\sum_{n=-m}^{m} g(1-n) = g(m) + g(m+1)$ and letting $m\to\infty$, the desired conclusion follows by the Squeezing Theorem. $\square$

Remarks.

  1. The condition 3 that $|g(x)| \to \infty$ is necessary. Indeed, if we consider $g(x) = \arctan x$, then all of the conditions 1, 2, 4 are satisfied but we can prove that $f(x)$ exists but is not identical as $x$ (although it is very close to $x$).

  2. The condition 4 that $g(x+a)/g(x) \to 1$ is also necessary. Indeed, for $g(x) = \sinh x$ we have $f(x) = \sinh x/\sinh 1$.

Sangchul Lee
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