1
$\begingroup$

Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that

$$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$

exists and compute the limit.

Attempt:

Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &=\int_0^1 f(0)dx=1. \end{align*}

Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.

My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified.

$\endgroup$
  • $\begingroup$ Your argument would be more rigorous if you separate the boundary point $1$. Since $\{1\}$ has measure $0$, it would not affect the final result. In addition, $f(x^n)$ is bounded by $g(x) = \sup_{0 \leq x \leq 1} f(x)$, which is finite number due to continuity and compactness of $[0, 1]$. “$f_n(x^n)$” is redundant. $\endgroup$ – Zhanxiong Jan 6 '18 at 21:50
1
$\begingroup$

Let $f_n(x) = f(x^n)$ note that the $f_n$ are uniformly bounded (since $f$ is countinuous on $[0,1]$) and $f_n(x) \to 1$ for all $x <1$. Then $\lim_n \int f_n = \int \lim_n f_n = 1$.

$\endgroup$
1
$\begingroup$

Yes it is clearly correst because $f$ is continuous so you are right. Further more, let $u_n$ be the general term of a sequence given by $$ u_n=f\left(x^n\right) $$ $x \in \left[0,1\right]$ so with $f$ continous, the sequence $\displaystyle \left(u_n\right)_{n \in \mathbb{N}}$ converges simply to $g$ defined by $$ g\left(x\right)=f\left(0\right) \text{ if }x \in \left[0,1\right[ \text{ and }f\left(1\right)\text{ for }x=1 $$ Then you apply dominated convergence theorem on $\displaystyle \left(u_n\right)_{n \in \mathbb{N}}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.