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Problem: Let $f,g\colon \mathbb{R} \to \mathbb{R}$ be functions such that $f$ is differentiable and for every $x,h \in \mathbb{R}$ one has $f(x+h)-f(x-h)=2hg(x)$. Prove that $f$ is a polynomial of degree at most $2$.

My Attempt: As $f(x)$ is differentiable with respect to $x$ and $2hg(x)=f(x+h)-f(x-h)$, we know that $g(x)$ is differentiable with respect to $x$. Moreover, we have $$ \begin{split} 2hg(x) &= f(x+h)-f(x-h) \\ 2hg(x) &= f(x+h)-f(x)+f(x)-f(x-h) \\ 2g(x) &= \frac{f(x+h)-f(x)}{h} - \frac{f(x-h)-f(x)}{h} \end{split} $$ so that as $h \to 0$ $$ 2g(x) =\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} - \frac{f(x-h)-f(x)}{h} = f'(x)-(-f'(x))=2f'(x) $$ This shows that $g(x)=f'(x)$. As $g(x)$ is differentiable, this shows that $f(x)$ is twice differentiable and that $g'(x)=f''(x)$. All that remains is to show that $f''(x)=g'(x)$ is constant.

However, I've tried many things and have yet to see why $g'(x)$ is constant. Any thoughts on how to do this?

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If $f\in C^3$ this follows immediately from triple-differentiating the first equation with respect to $h$ and evaluating as $h\to 0$. So the heart of question is to show this is still true when $f$ is merely differentiable. – user7530 Jan 10 at 20:48
@user7530 why $3$? Wouldn't $2$ be enough? (maybe I'm overlooking it) – Silvia Ghinassi Jan 10 at 20:51
@SilviaGhinassi I need to differentiate three times to get $f'''(x) = 0$ (and hence $f$ is quadratic). – user7530 Jan 10 at 20:53
Oh, I think differentiating two times you can conclude $f''$ is constant, which is the same as saying what you are saying. I haven't written anything down anyway, so don't believe me. – Silvia Ghinassi Jan 10 at 20:54
OP did that, if I am not mistaken. – Silvia Ghinassi Jan 10 at 21:01
up vote 11 down vote accepted

You have concluded $f$ is twice differentiable. Once we have that, we can differentiate the first equation two times with respect to $h$, to get $$ f''(x+h)-f''(x-h)=0. $$ Since this holds for every $x$ and $h$ we get that $f''$ is constant.

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How does this imply that $f''$ need be constant? – Kyle L Jan 10 at 21:14
Nevermind! $f$ is trice differentiable. Thanks! – Kyle L Jan 10 at 21:22
@KyleL for every $y,z\in\mathbb{R}$, you can write $x=(y+z)/2$ and $h=(y-z)/2$. Then $f''(y)=f''(x+h)=f''(x-h)=f''(z)$. – JonSK Jan 10 at 21:23
@SilviaGhinassi , What was the motivation behind differentiating it with respect to $h$ and not wrt. $x$? – Cupid Jan 11 at 8:59
@MosBlack We dont yet know what the second derivative of 2hg(x) wrt x is. However g(x) is constant wrt h. – Taemyr Jan 11 at 10:20

Differentiate $f(x+h)-f(x-h)=2hg(x)$ with respect to $h$ to get $$ f'(x+h)+f'(x-h)=2g(x). $$ Now set $h=0$ to deduce that $f'(x)=g(x)$ for all $x$. Thus the (continuous) function $g=f'$ has the "harmonic function" property $$ g(x) ={g(x+h)+g(x-h)\over 2},\qquad\forall x,h. $$ This is enough (by continuity of $g$) to imply that $g$ affine ($g(x) = mx+b$) and so $f$ must be quadratic.

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For arbitrary $x$ and $h$ we have

$$\eqalign{ g(x-h)+g(x+h)&=\frac1{2h}(f(x+2h)-f(x))+\frac1{2h}(f(x)-f(x-2h))\\ &=\frac2{4h}(f(x+2h)-f(x-2h))\\ &=2g(x)\\ g(x+h)-g(x)&=g(x)-g(x-h) }$$

By induction, for all integers $n,$


Now for arbitrary $y,$ apply this identity with $x=0$ and $h=y/n$:

$$g(y)=g(0)+y\frac{g(\frac y n)-g(0)}{\frac yn}$$

and the right hand side converges to $g(0)+yg'(0)$ as $n\to\infty.$ Thus $g$ is a polynomial of degree $1$.

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This is a nice solution, but the last line is a tad confusing. I assume you mean that $y$ is fixed and that $h = y/n$, so $h \to 0$ as $n \to \infty$. – Jacob Maibach Jan 11 at 4:43
That's it. It have rephrased it slightly hoping to make this clearer. – Justpassingby Jan 11 at 8:23

We use your result: Namely, that $f$ is twice differentiable.

Let $x \in \mathbb{R}$ be arbitrary, fixed. Define the function $\xi: \mathbb{R} \rightarrow \mathbb{R}$ as $\xi(h)=f(x+h)$. Then the equality says $$\xi(h)-\xi(-h)=2hg(x).$$ Differentiating twice, we get $$\xi''(h)-\xi''(-h)=0$$ By the chain rule, it follows that $f''(x+h)-f''(x-h)=0$ for every $x,h$.

Now, suppose $f''$ is not constant. Then, there exists $\alpha, \beta$ such that $f''(\alpha)\neq f''(\beta) \implies f''(\alpha)-f''(\beta) \neq 0$. Taking $x=\frac{\alpha+\beta}{2}$ and $h=\frac{\alpha-\beta}{2}$ yields a contradiction with the previous conclusion.

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