# $f$ is T periodic and $f(x) + f'(x) \ge 0 \Rightarrow f(x) \ge 0$

Let $$f: \Bbb R \to \Bbb R$$ be a function such that $$f'(x)$$ exists and is continuous over $$\Bbb R$$. Moreover, let there be a $$T > 0$$ such that $$f(x + T) = f(x)$$ for all $$x \in \Bbb R$$ and let $$f(x) + f'(x)\ge 0$$ for all $$x \in \Bbb R$$.

Show that $$f(x) \ge 0$$ for all $$x \in \Bbb R$$.

My attempt: $$f(x) \ge 0 \iff f(x) \ge f'(x) - f'(x) \iff f(x) + f'(x) \ge f'(x)$$.

Thus, it is enoguh to show that $$0 \ge f'(x)$$.

$$\iff 0 \ge \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$$

I do not know how to proceed from here. I know that $$f'$$ also has a periodicity of T but I do not know how to use that here.

Am I on the right track? How can I use the periodicity of $$f$$ to solve the problem?

Let $g(x)=\mathrm e^xf(x)$, then $g'(x)=\mathrm e^x(f'(x)+f(x))\geqslant0$ hence: $$(1)\ \textit{The function g is nondecreasing.}$$ Since $f$ is continuous and periodic, $f$ is bounded, say $|f(x)|\leqslant C$ for every $x$, hence $|g(x)|\leqslant C\mathrm e^x\to0$ when $x\to-\infty$, that is: $$(2)\ \textit{The function g has limit 0 at -\infty.}$$ Properties (1) and (2) of $g$ imply together that $g\geqslant0$ everywhere, hence $f\geqslant0$ everywhere.

Examples: Consider $$f(x)=c\,\mathrm e^{w\cos(ux+v)},$$ for every $c\geqslant0$, $u\ne0$, $|uw|\leqslant1$ and $v$, then $f$ has period $2\pi/|u|$ and, for every $x$, $$f'(x)+f(x)=(1-wu\sin(ux+v))\,f(x)\geqslant0.$$ For example, if $c=w=u=1$, $v=0$, one gets the function $f$:

$\qquad\qquad\qquad$

...And the function $f'+f$:

$\qquad\qquad\qquad$

We don't even need that $$f'(x)$$ is continuous! It is enough to show that $$f(x) \ge 0$$ on $$[0,T]$$, because of periodicity. Since $$[0,T]$$ is compact, $$f$$ has a minimum on $$[0,T]$$. Let $$x_0 \in [0,T]$$ be such that $$f(x_0)$$ is minimal. If $$x_0 =0$$ or $$x_0 =T$$, then $$f(x_0)$$ is also minimal in $$[-T,2T]$$, because of periodicity.

A necessary condition for a minimum is that $$f'(x_0) =0$$. (And it is also valid in the case $$x_0 =0$$ or $$x_0 =T$$, because of the above remark.) Thus we get $$f(x_0) = f(x_0)+f'(x_0) \ge 0$$. This shows that $$f(x) \ge 0$$ everywhere.

• You do need $f'(x)$ continuous when you say "a necessary condition for a minimum is that $f'(x_0)=0$". Nov 30, 2018 at 13:41
• That is not true! For example, if $x_0$ is a minimum, then we have $f(x_0) \le f(x)$ for $|x-x_0| \le \delta$. Thus $f'(x_0) =\lim_{h \downarrow 0} (f(x_0+h)-f(x_0))/h \le 0$. On the other hand $f'(x_0) =\lim_{h \downarrow 0} (f(x_0-h)-f(x_0))(-h) \ge 0$. Thus $f'(x_0)=0$. In this (almost trivial) proof I haven't used coninuity of $f'$ and we don't need it! Nov 30, 2018 at 13:44
• Yeah, I was thinking about $f(x)=|x|$, but in that case $f'(0)$ doesn't even exist. Dec 1, 2018 at 10:21

Since $f$ is periodic, it has maximum and minimum. Choose the period $[a,b]$, such that $f(a)=f(b)=M$, where $M$ is the maximum.

It can be seen that, there is a point $f(c)=m$, where $m$ is the minimum. c is golbal minimum, and thus local minimum, so:

$$f'(c)=0$$

$$f+f'\ge 0$$

we see

$$f(c)\ge 0$$. Since $f(c)$ is the minimum, we are done!

Consider such $f(x)$ to be non-constant. There must be a global minimum within each period, say at $x=x_0$. Since $f(x)$ is differentiable and continuous, $f'(x_0) = 0$. Hence $$f(x) \ge f(x_0) = f(x_0)+f'(x_0) \ge 0$$

suppose $$f(x)$$ is negative for all values of $$x$$. Periodicity tells us that there must be solutions to $$f'(x)=0$$ and for such $$x$$ we would then have $$f(x)+f'(x)=f(x)<0$$.

Suppose that $$f(x)$$ is negative for some values of $$x$$ but not all. Let $$a,b$$ be zeroes of $$f(x)$$ such that $$a. Specifically, let $$x_0$$ be some value for which $$f(x_0)<0$$ and let $$a$$ be the greatest upper bound of $$\{x\,:\,x and $$b$$ defined similarly on the other side. Since $$f(a)=0=f(b)$$ there is some $$c$$ with $$a and $$f'(c)=0$$. But for that value we must have $$f(c)+f'(c)=f(c)<0$$ contrary to assumption.

• Nice application of Rolle's Theorem. Nov 30, 2018 at 12:30

Say a period is $[a,a+T]$. Since $f$ is continuous, it attains a minimum on that interval, say at $c$. (If you have $c = a$, then change the period to $[a - T/2,a + T/2]$ so that $c$ becomes an interior point.)

We must have $f'(c) = 0$ there, so $f(c) \geq 0$. So the minimum value of $f$ is nonnegative.

$f \ne 0$

$f(x)=0 \Rightarrow f'(x) \ge 0$, hence $f$ can cross the X-axis at most once. periodicity means that it cannot cross at all.

if $f$ is non-positive then $f' \ge 0$ so $f$ is monotone increasing. again this contradicts periodicity

• Why is this downvoted? This answer is essentially saying if $f(x)<0$ somewhere, then $f(x)$ is strictly increasing there, contradicting the fact that $f(x)$ is periodic. Nov 23, 2014 at 16:48