Find $\lim_{n\to\infty}\frac{g(t+n)}n$ for $g(t)=\int_0^tf(x)\,dx$, where $f(x+1)=f(x)$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(x+1)=f(x) \quad \forall x\in\mathbb{R}$.

Define $g(t)=\displaystyle\int_0^tf(x)\,dx$, $t\in\mathbb{R}$ and $h(t)=\displaystyle\lim_{n\to\infty}\frac{g(t+n)}{n}$ (provided it exists).

Show that $h(t)$ is defined $\forall t\in \mathbb{R}$ and is independent of $t$.

As the period of $f$ is $1$, we can say that $\displaystyle\int_0^tf(x)\,dx=t\displaystyle\int_0^1f(x)\,dx$. But what about $\displaystyle\int_0^{t+n}f(x)\,dx$? It is also seen that $g'(t)=f(t)$. How do I find the limit?

Answer 1

As the period of $f$ is $1$, we can say that $\displaystyle\int_0^tf(x)\,dx=t\displaystyle\int_0^1f(x)\,dx$

Is false.

For example, take $f(x)=\sin(2\pi x)$. Then

$$t\cdot\int_0^1 f(x)dx=0=t\cdot 0 = 0$$

while $$\int_0^t f(x)dx$$

is not $0$ for all values of $t$.

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To solve the problem, first define the constant

$$A=\int_0^1 f(x)dx$$

Then, you can easily see that

$$g(t)=A\cdot \lfloor t\rfloor + \int_0^{\{t\}} f(x)dx$$

where $\{t\}$ is the fractional part of $t$ and $\lfloor t\rfloor$ is the largest integer smaller than $t$.

Once you have $g$ in that shape, both tasks should be simple to solve.

Answer 2

For a fixed $t$,

$$g(t+n) = \int_0^{t+n} f(x) dx = \int_0^{t} f(x) dx + \int_t^{t+n} f(x) dx = \int_0^{t} f(x) dx +\sum_{i=0}^{n-1} \int_{t+i}^{t+i+1} f(x) dx$$

Lemma: if a continuous function $f$ has period $T$, then for any $a$, $\int_{a}^{a+T} f(x) dx = \int_0^T f(x)dx$

Proof: Let $F$ denote a primitive of $f$ and let $g:x\to F(x+T)-F(x)$. Note that $g'=0$ hence $g$ is constant. Note also that $\displaystyle \int_{a}^{a+T} f(x) dx = F(a+t)-F(a) = g(a) = g(0)=\int_0^T f(x)dx$

Hence $\displaystyle g(t+n) = \int_0^{t} f(x) dx +\sum_{i=0}^{n-1} \int_{0}^{1} f(x) dx = \int_0^{t} f(x) dx + n \int_{0}^{1} f(x) dx$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With Stolz-Ces$\mathrm{\grave{a}}$ro Theorem:

\begin{align} \color{#f00}{\mathrm{h}\pars{t}} & = \lim_{n \to \infty}{\mathrm{g}\pars{t + n} \over n} = \lim_{n \to \infty} {\mathrm{g}\pars{t + n + 1} - \mathrm{g}\pars{t + n} \over \pars{n + 1} - n} = \lim_{n \to \infty}\int_{t + n}^{t + n + 1}\mathrm{f}\pars{x}\,\dd x \\[3mm] & = \lim_{n \to \infty}\int_{t}^{t + 1}\mathrm{f}\pars{x + n}\,\dd x = \lim_{n \to \infty}\int_{t}^{t + 1}\mathrm{f}\pars{x}\,\dd x = \color{#f00}{\int_{t}^{t + 1}\mathrm{f}\pars{x}\,\dd x} \end{align}

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Moreover, \begin{align} \mathrm{h}'\pars{t} & = \totald{}{t}\int_{t}^{t + 1}\mathrm{f}\pars{x}\,\dd x = \mathrm{f}\pars{t + 1} - \mathrm{f}\pars{t} `= 0\quad\imp\quad \color{#f00}{\mathrm{h}\pars{t}}\ \mbox{is}\ t\mbox{-independent} \\[3mm] &\ \mbox{It means}\ \,\mathrm{h}\pars{t} = c = \mbox{constant. In particular,}\ c = \,\mathrm{h}\pars{t} = \color{#f00}{\int_{-1/2}^{1/2}\mathrm{f}\pars{x}\,\dd x} \\[3mm] & \mbox{Indeed,}\ c = \color{#f00}{\,\mathrm{h}\pars{t}} = \color{#f00}{\int_{a}^{a + 1}\mathrm{f}\pars{x}\,\dd x} = \pars{~\mbox{a constant}~}\,,\ \forall\ a\ \in\ \mathbb{R} \end{align}