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Let $f\in C^4(\mathbb{R})$ and satisfy $$f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x+\theta h)h^2,$$ with $\theta\in (0,1)$ is a constant independent of $x,h$. Show that $f$ is a polynomial with degree at most $3$.

I have no idea on it...

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Differentiate the equation with respect to $x$ twice and then substitute $h$ by $\theta h$, we obtain

$$f''(x+\theta h)=f''(x)+f^{(3)}(x)\theta h+\frac12f^{(4)}(x+\theta^2 h)\theta^2 h^2$$

Substitute this back into original equation, we get

$$f(x+h) = f(x) + f'(x)h+\frac{h^2}{2}\left[f''(x)+f^{(3)}(x)\theta h+\frac12f^{(4)}(x+\theta^2 h)\theta^2 h^2\right]$$

Since $f\in C^{4}(\mathbb{R})$, we have a Taylor expansion of $f(x+h)$ of the form

$$f(x+h) = f(x) + f'(x)h+\frac12 f''(x)h^2 + \frac{1}{3!} f^{(3)}(x)h^3 + \frac{1}{4!}f^{(4)}(x + \phi_{x,h} h) h^4$$ for some $\phi_{x,h} \in (0,1)$ which in general depends on both $x$ and $h$. Compare these two expansion, we get

$$\theta f^{(3)}(x) + \frac12 f^{(4)}(x+\theta^2 h)\theta^2 h = \frac13f^{(3)}(x) + \frac{1}{12} f^{(4)}(x + \phi_{x,h} h)h$$

If $f^{(3)}(x)$ is not identically zero (i.e $f(x)$ not a polynomial of degree $2$), then we can take limit of $h \to 0$ at suitable $x$ to conclude $\theta = \frac13$. This leads to the identity $$\frac{1}{18} f^{(4)}(x + \frac19 h) = \frac{1}{12} f^{(4)}(x + \phi_{x,h} h)$$ Since $f \in C^4(\mathbb{R})$, we can then take limit $h \to 0$ again to get

$$\frac{1}{18} f^{(4)}(x) = \frac{1}{12} f^{(4)}(x) \quad\implies\quad f^{(4)}(x) = 0 \quad\text{ for all } x$$ and hence $f(x)$ is a polynomial with degree at most $3$.

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(+1) Thats Awesome !! Can this method be extended to the case where $f\in C^{n+2}$, and $f(x+h)=f(x)+hf'(x)+\ldots+\frac{h^n}{n!}f(x+\theta h)$ for some constant $\theta$, where we can show $\theta = \frac{1}{n+1}$ and finally $f$ is a polynomial ? – r9m Mar 20 '14 at 9:48
@r9m, Yes, you can. If $f \in C^{n}$ (no need to assume in $C^{n+2}$ ) and satisfies the delayed ODE for fixed $\theta$, then we can use the delayed ODE to show $f \in C^{\infty}$. All the construction above essentially goes through w/o too much change and $f$ is a polynomial of degree at most $n+1$. – achille hui Mar 20 '14 at 10:09

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