Cauchy functional equation and the implicit function theorem

Question

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 $\eta$.

Answer 1

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!

Answer 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

$$h(x)=\int_{x}^{x+\xi}f'(y)\text{d}y-\int_{0}^{\xi}f'(y)\text{d}y-f(0)$$

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.

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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

$$h(x)=f'(z_x)\xi-f(\xi)$$

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

$$\lim_{x\to\pm\infty}f'(z_x)=\pm\infty$$

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