Showing a function is a polynomial

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $n,$

$$g(x+nh)-g(x)=n(g(x+h)-g(x))$$

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 $1$.

Answer 4

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.