The problem statement:

Suppose $f:\mathbb{R} \to \mathbb{R}$ is continuously differentiable, $f'(x)$ is strictly increasing, with $\lim_{x \to -\infty}f'(x) = -\infty$, $\lim_{x \to \infty}f'(x) = \infty$, $f(0) \neq 0$.

a) Prove that $\forall \xi \neq 0$, there exists an $\eta$ such that $f(\xi + \eta) = f(\xi) + f(\eta)$

b) Prove that through this point $(\xi, \eta)$ there is a solution $y=\phi(x)$ of $f(x+y) = f(x) + f(y)$ which is unique in a neighborhood of $(\xi, \eta)$.

c) Construct an example to show that when $\xi = 0$ there may be no corresponding $\eta$.

  • $\begingroup$ You've used the name $f$ in several places that, to me, seem to be incompatible. Can you clarify a bit? $\endgroup$
    – Ian
    Jan 2, 2015 at 4:38
  • $\begingroup$ What is it that seems incompatible, the question or the answer? The question is copied verbatim from an old exam, but it is certainly possible I did something wrong in my answer. $\endgroup$
    – dnk_js91
    Jan 2, 2015 at 4:42
  • $\begingroup$ You introduce $f$ at the top and then write the functional equation in terms of $f$. Are these the same $f$? $\endgroup$
    – Ian
    Jan 2, 2015 at 4:49
  • 2
    $\begingroup$ Yes, any reference to $f$ throughout is to the definition given in the problem statement. $\endgroup$
    – dnk_js91
    Jan 2, 2015 at 4:52
  • 1
    $\begingroup$ The argument you gave is good for part (b) but not for part (a). The implicit function theorem needs the starting point that is on the graph of the function. Instead consider $h(x):=f(x+\xi)-f(x)-f(\xi)$ and use the information about the limits. $\endgroup$
    – Pp..
    Jan 2, 2015 at 4:55

2 Answers 2


To finish (a) let us use the $$h(x):=f(x+\xi)-f(x)-f(\xi).$$

We have $f(x)=\int_{0}^{x}f'(y)\text{d}y+f(0)$ so


The sum of the last two terms is a constant. The first term is an integral over an interval of constant length. Now use the given limits to show that if you move $x\to-\infty$ the first term goes to $-\infty$ (this is because the integrand goes to $-\infty$). Similarly the first integral goes to $+\infty$ when you move $x\to+\infty$.

Now apply intermediate value theorem.

The other parts you already know how to do.

You wanted to use the mean value theorem. In that case it gives us, for each $x$, a $z_x\in[x,x+\xi]$ such that $f'(z_x)\xi=f(x+\xi)-f(x)$. Then


Notice now, that when you move $x$ towards $-\infty$ or towards $\infty$ then $z_x$ gets pushed towards there as well. So,


Now, as in the other proof, apply the intermediate value theorem to $h(x)$.

  • $\begingroup$ Great, thank you so much! I've collected my entire answer into one post for future reference but I appreciate all the help! $\endgroup$
    – dnk_js91
    Jan 2, 2015 at 5:51

a) Define $h(x) = f(x+\xi) - f(x) - f(\xi)$. Applying the mean value theorem to $f(x+\xi)-f(x) = \xi f'(\gamma)$, where $\gamma \in (x, x+\xi)$. If we take the limit as $x \to \pm \infty$, we must have that $\gamma$ tends to $\pm \infty$ as well. This implies $\lim_{x \to -\infty} h(x) = -\infty$ and $\lim_{x \to \infty} h(x) = \infty$. Since $h$ is continuous by construction, the intermediate value theorem gives $h(\eta) = 0$ for some $\eta$.

b) Let $g(x,y) = f(x+y) - f(x) - f(y)$. By a), we know that $\exists (\xi, \eta)$ s.t. $g(\xi, \eta) = 0$. Further we note that the partial derivatives of $g$ are sign definite, since $$g_x = f'(x+y) - f(x) \quad g_y = f'(x+y) - f'(y)$$ and we can conclude these are nonzero by the fact that $f'$ is one-to-one on all of $\mathbb{R}$ (continuous monotone function is one-to-one). Then by the implicit function theorem, there exists a solution $y=\phi(x)$ to the functional equation that is unique in a neighborhood of $(\xi, \eta)$.

c) Consider $f(x) = x^2 + 1$. $f'(x)$ satisfies the requirement, but $$x^2+1=f(x) = f(x + 0) \neq f(x) + f(0) = x^2+2$$ Clearly we cannot apply the implicit function theorem because we do not have the condition in a). We can also deduce that we must have $f(0) = 0$ which is a contradiction to the problem statement.

Many thanks to Pp.. for nursing me through this one!


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