# With the condition $\lim_{x\to \infty}(f(x+a)-f(x))=0$, how to construct $h(x)$?

Assume $f(x)\in C[0,+\infty)$，and for all $a\geqslant 0$, we have \begin{align*} \lim_{x\to \infty}(f(x+a)-f(x))=0 \tag{*}. \end{align*} Prove that there exists $g(x)\in C[0,+\infty)$ and $h(x)\in C^1[0,+\infty)$ such that $f(x)=g(x)+h(x)$, and such that they satisfy \begin{align*} \lim_{x\to \infty}g(x)=0,~~\lim_{x\to \infty}h'(x)=0. \end{align*}

My thought is let $h(x)=\frac1 a\int_x^{x+a}f(t)\,dt$, then it is easy to see $\lim_{x\to \infty}h'(x)=0$, but I can't explain that $\lim_{x\to \infty}g(x)=\lim_{x\to \infty}f(x)-h(x)=0$. It seems we should try proving $\lim_{x\to +\infty}f(x)$ exists by using the condition of (*), but I'm not sure whether it's true or false.

• If $f(x) = \ln(x+1)$ then (*) is true for all $a \ge 0$ but $\lim_{x\to\infty} f(x)$ does not exist. Apr 11, 2017 at 0:08
• @DanielSchepler Is that a problem ? If $f(x) = \ln(x+1)$, you can simply take $g = 0$ and $h=f$. Apr 11, 2017 at 0:21
• @JoelCohen I was just responding to your conjecture that $\lim_{x\to \infty} f(x)$ exists. Apr 11, 2017 at 0:22
• @DanielSchepler A good counterexample！ Apr 11, 2017 at 0:29
• Another example to consider would be $f(x) = \frac{1}{x} \sin(x^2)$ -- there $\lim_{x\to\infty} f'(x)$ does not exist. Apr 11, 2017 at 0:35

EDIT : I'm assuming uniform continuity here, so my proof only works in that case.

Let's chose $a = 1$ and set $h(x) = \int_x^{x+1} f(t) \, dt$. We begin by writing

$$h(x)-f(x) = \int_0^{1} (f(x+t)-f(x)) \ dt$$

The integrand converges to $0$ pointwise (from condition (*)), but this is not quite sufficient ! We'll have to be a bit more careful and also use the uniform continuity of $f$.

Let $\epsilon > 0$. Because $f$ is uniformly continuous, there exists an integer $n > 0$ such that for all $x,y \ge 0]$ with $|x-y|\le \frac{1}{n}$, we have $|f(x)-f(y)| \le \epsilon$.

Now we use condition (*) to get that for $1 \le k \le n$, there exists $x_k$ such that for all $x \ge x_k$,

$$\left|f\left(x+\frac{k}{n}\right)-f(x)\right| \le \epsilon$$

We set $x_0 = max_{1 \le k \le n}(x_k)$. Now, for $x \ge x_0$ we have

$$\begin{eqnarray*} |h(x)-f(x)| &=& \left| \int_0^{1} (f(x+t)-f(x)) \ dt \right|\\ &\le& \int_0^{1} |f(x+t)-f(x)| \ dt \\ &\le& \sum_{k= 1}^{n} \int_{\frac{k-1}{n}}^{\frac{k}{n}} \underbrace{\left|f(x+t)-f\left(x+\frac{k}{n}\right)\right|}_{\le \epsilon \ \text{ (from continuity)}} + \underbrace{\left|f\left(x+\frac{k}{n}\right)-f(x)\right|}_{\le \epsilon \ \text{ (from (*))}} \ dt \\ &\le& 2 \epsilon \end{eqnarray*}$$

• Thanks,It seems in your solution that you suppose f is uniformly continuous. Apr 11, 2017 at 2:55
• @mbfkk indeed, typed this a bit too fast. I edited my answer to reflect this restriction. Not sure there's a way to mend this current proof without any way to uniformly control the integrand (either uniform continuity, or an uniform version of condition (*)). Apr 11, 2017 at 12:07
• We can use Baire Category Theorem to prove $f(x)$ is uniformly continuous,there is the answer. With the condition $\lim{x \to \infty}(f(x+a)−f(x))=0$, how to prove that $f(x)$ is uniformly continuous?
– ling
May 5, 2019 at 2:25