# Show that solutions of $f'(x)=f(x+\frac{\pi}{2})$ have real roots.

Given that $$f: \mathbb{R} \to \mathbb{R}$$ is differentiable and $$f'(x)=f(x+\frac{\pi}{2})$$ for all $$x\in \mathbb{R}$$, prove that there exists $$x_0$$ such that $$f(x_0)=0$$

I thought that $$f(x)=\cos(x)$$ but I don't know how to prove it. Moreover is there a solution without using the $$\cos(x)$$?

• $f(x)$ could be of the form $a\sin x+b\cos x$ – Shubham Johri Feb 20 at 9:57
• Stop trying to figure out what $f$ is, and instead think about what you can prove from what you already know about $f$. If $f(x) \ne 0$ anywhere, what would that say about $f$? (Note that $f$ is continuous.). In turn, what does that say about $f'$, given their relationship? And what does that say about $f$ again? Can these two properties of $f$ be made compatible with $f'(x) = f(x + \pi/2)$? – Paul Sinclair Feb 20 at 21:34
• @PaulSinclair I see. if $f$ has no zero then $f$ has to be either strictly positive, or strictly negative. Say that it is strictly positive, this means also that $f'$ is strictly positive. So $f$ should be strictly increasing. could I come up with a contradiction from here? – karhas Feb 21 at 19:29
• @ShubhamJohri could you describe a way to see this? Or provide me with some reference? – karhas Feb 21 at 19:31

Let us consider the equation $$f'(x) = f(x+a)$$ with $$a \ge 1$$ instead. Assume that $$f$$ has no zeros. Without loss of generality we can assume that $$f$$ is strictly positive.

It follows that $$f$$ and $$f'$$ are strictly increasing, so that for $$x < y$$ $$\tag {*} f(y) > f(x) + (y-x) f'(x) \, .$$ In particular, $$\lim_{x \to \infty} f(x) = +\infty$$.

On the other hand, using $$(*)$$ again, $$f'(x) = f(x + a) > f(x) + a f'(x)$$ implies $$(a-1)f'(x) + f(x) < 0$$

If $$a=1$$ then $$f(x) < 0$$ gives an immediate contradiction. If $$a > 1$$ then the last inequality shows that $$e^{x/(a-1)} f(x)$$ is decreasing, so that $$f(x) \le f(0) e^{-x/(a-1)} \, ,$$ contradicting the fact that $$\lim_{x \to \infty} f(x) = +\infty$$.

Remark: If $$0 < a \le 1/e$$ then $$e^{\lambda a} = \lambda$$ has a real solution $$\lambda$$, and $$f(x) = e^{\lambda x}$$ satisfies the functional equation and has no zeros (compare How to solve differential equations of the form $f'(x) = f(x + a)$).

• Bravo and interesting. I was trying to find a proof for the more general equation $f^\prime(x) = f(x+a)$ with $a \gt 0$. What you did can be generalized for $a \gt 1$. The result is also true when $a=1$. What about $a \lt 1$? I don't know for the moment! – mathcounterexamples.net 2 days ago
• @mathcounterexamples.net: It is definitely wrong for small $a$. If $e^{\lambda a} = \lambda$ has a real solution $\lambda$ then $f(x) = e^{\lambda x}$ satisfies the given equation. Accordig to math.stackexchange.com/a/61889, this works for $a \le 1/e$. – Martin R 2 days ago