Calculating functional sequence

Question

Let $A\in \mathbb{R}$ and function f is continous on $\left [ 0,\infty \right ]$ such as $\lim_{n \to \infty }f(x)=A$.

Calculate:

$\lim_{n \to \infty } \int_{0}^{1}f\left ( nx \right )dx$

and find at least two approaches for solving this problem.

1. My idea as the first aprroach is to use supremum criterion.

For this $\lim_{n \to \infty } \int_{0}^{1}f\left ( nx \right )dx=0$, we can use substitution, so that:$\lim_{n \to \infty } \frac{1}{n} \int_{0}^{u}f\left ( u \right )du=0$

But then I do not know, how to continue. Can anyone help me?

Answer 1

\begin{align*} \int_{0}^{1}f(nx)dx=\dfrac{1}{n}\int_{0}^{n}f(x)dx. \end{align*} Given $\epsilon>0$, there is a $M>0$ such that $|f(x)-A|<\epsilon$ for $x\geq M$, then \begin{align*} \left|\dfrac{1}{n}\int_{0}^{n}f(x)dx-A\right|&=\left|\dfrac{1}{n}\int_{0}^{n}(f(x)-A)dx\right|\\ &\leq\dfrac{1}{n}\int_{0}^{M}|f(x)-A|dx+\dfrac{1}{n}\int_{M}^{n}|f(x)-A|dx\\ &\leq\dfrac{1}{n}\int_{0}^{M}|f(x)-A|dx+\dfrac{1}{n}(n-M)\epsilon, \end{align*} so \begin{align*} \limsup_{n\rightarrow\infty}\left|\dfrac{1}{n}\int_{0}^{n}f(x)dx-A\right|\leq 0+\epsilon. \end{align*}

Answer 2

hint

By the substitution, $nx=t$, the integral becomes $$A_n=\frac 1n\int_0^nf(t)dt$$

but for great enough $t$, let say for $t\ge T$, we have

$$-\epsilon <f(t)-A<\epsilon$$ thus for large $n,$

$$A_n -A=\frac 1n \int_0^T(f(t)-A)dt +\frac 1n\int_T^n(f(t)-A)dt$$