Let $f: \mathbb{R} → \mathbb{R} $ be a function such that $f(x)$ is differentiable on all $\mathbb{R}$ and $\lim_{x\to \infty}(f(x)-f(-x))=0$.
Prove there exists $x_{0} \in \mathbb{R}$ such that $f'(x_{0})=0$
I tried proving it by contradiction using that the limits at infinity and minus infinity are equal I'm having hard time formalizing it.
Any help appreciated.
Either $f$ has a local extremum somewhere at $x_0\in\mathbb R$ (a maximum or minimum), either it is monotone. In the first case, then necessarily $f'(x_0)=0$, in the second case, the fact that $\lim_{x\to+\infty}(f(x)-f(-x))=0$ implies that $f$ is constant, and thus $f'=0$ everywhere.
Hint: Assuming $f'$ is continuous and integrable you could use the fundamental theorem of calculus.
Let the function tend to value L for x tending to positive Infinity and the function has to tend towards same value for x tending to negative Infinity. If The convergence to this value is not monotonic, or oscillatory then it has to have a Maxima or minima. If the convergence is monotonic, and from the top side (values of function is greater than limit) or bottom side ( values of function is smaller than limit) both sides, then apply Rolle's theorem. And if the convergence is not in same direction both sides, then there will be atleast two stationary points using Rolle's theorem.
finally if there is no converging limit, as in the case of $y=x^2$, in those cases the functions have to have positive and negative derivatives and one can use intermediate value theorem for derivative function.
As $f$ is continuous on $\mathbb{R}$, it is so on $[0,1]$. Let $\alpha=\sup_{x\in [0,1]} f(x)$ and $\beta=\inf_{x\in [0,1]} f(x)$, then by continuity there exists some $x_0,x_1$ such that $f(x_0)=\beta$, $f(x_1)=\alpha$. If any of $x_0$ or $x_1$ lies inside $[0,1]$ (i.e. not an endpoint) then $f'$ vanish at that point, since we'd have local maxima/infima.
So assume that $x_0,x_1$ are endpoints, and WLOG we may assume that $x_1=1$ (otherwise consider $f(-x)$ instead). Further, we may assume $f'$ does not vanish at endpoints.
If $\alpha=\beta$ then $f$ is constant on $[0,1]$ so we are done. Assume $\alpha>\beta$.
Now assume for a contradiction that $f$ is injective. It follows that $f(\mathbb{R}\setminus [0,1])\subseteq \mathbb{R}\setminus [\beta,\alpha]$. By intermediate value theore, (or by continuity + connectedness, whichever way you'd like) it then follows that $f(1,\infty)$ and $f(-\infty,0)$ are connected, ie intervals. Therefore, either $f(1,\infty)\subset (\alpha,\infty)$ or $f(1,\infty)\subset(-\infty,\beta)$. But second case cannot be possible, since $\lim_{x\to 1+}f(x)=\alpha$ by continuity, which cannot happen if $f(1,\infty)\subset(-\infty,\beta)$.
Simiarly, we have $f(-\infty,0)\subset (-\infty,\beta)$. But then for large enough $x$ (>1) we have that $f(x)-f(-x)>\alpha-\beta>0$, contradictory to assumption. So $f$ is non-injective, and there exist $x_1,x_2$ distinct and $f(x_1)=f(x_2)$. Now conclude by Rolle's theorem.
Since $\lim_{x\to \infty}(f(x)-f(-x))=0$,
and $f(x)$ is differentiable and assumed continuous everywhere, by the Intermediate Value Theorem, there exists a $x_{0}$ such that:
$$f^{'}(x_{0})=\frac{\lim_{x\to \infty}f(x)-\lim_{x\to \infty}f(-x)}{x+x}=0.$$ (proved)