Let $d(x)=min_{n \in \mathbb Z}|x-n|$, Prove that $f(x)=\sum_{n=1}^{\infty} \frac{d(10^{n}x)}{10^{n}}$ is a continuous function on $\mathbb R$.

Question

Let $d(x)=min_{n \in \mathbb Z}|x-n|$, Prove that $f(x)=\sum_{n=1}^{\infty} \frac{d(10^{n}x)}{10^{n}}$ is a continuous function on $\mathbb R$.

b)Compute explicitly the value of $\int_{0}^{1}f(x)dx$.

I think in this question firstly I need to show that this series is uniformly convergent, then the limit $f(x)$ is continuous. Can anyone help me how I start this question or any other hint?

Answer 1

The series $\sum_{n=1}^{\infty}\frac{d(10^n x)}{10^n}\leq \sum_{n=1}^{\infty}\frac{1}{10^n} <\infty$ hence it is uniformly convergent, thus you get continuity in the limit.

For the integral,

$\int_0^1 f(x) dx = \sum_{n=1}^{\infty}\frac{1}{10^n} \int_0^1 d(10^n x)dx$.

Take $u = 10^n x$.

Inner integral becomes $\int_0^{10^n} 10^{-n} d(u) du$.

Now, for any $u$ whos fractional part lies in $[0,\frac{1}{2}]$, $d(u)=u$. For $u$ with fractional part in $(\frac{1}{2},1]$, $d(u)=1-u$.

Thus $10^{-n} \int_0^{10^n} d(u)du = 10^{-n}*10^{n} [\int_0^{\frac{1}{2}}u du + \int_{\frac{1}{2}}^{1}(1-u) du] = \frac{1}{4}$.

hence the overall integral is $\sum_{n=1}^{\infty}\frac{1}{4*10^n} = \frac{1}{36}$