Find $\lim\limits_{n\rightarrow\infty}\int\limits_0^1f(x^n)dx$

Question

Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function.

1. Show that for each $\epsilon\in(0,1)$, $\lim\limits_{n\rightarrow\infty}\int\limits_0^{1-\epsilon}f(x^n)dx=(1-\epsilon)f(0)$
2. Find $\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{1}f(x^n)dx$.
   Hint: Start by explaining why $f$ is bounded.

For my Answer:,
I have done the part one by using the fact that \begin{align*} \left|\int\limits_{0}^{1-\epsilon}f(x^n)dx-(1-\epsilon)f(0)\right|\leq \int\limits_0^{1-\epsilon}|f(x^n)-f(0)|dx \end{align*} And with the continuity of $f$ at zero along with $x^n\leq(1-\epsilon)^n\rightarrow0$
But for the part two what I can see is that: \begin{align*} \lim\limits_{n\rightarrow\infty}\int\limits_{0}^{1}f(x^n)dx &=\lim\limits_{n\rightarrow\infty}\lim\limits_{\epsilon\rightarrow0}\int\limits_{0}^{1-\epsilon}f(x^n)dx \end{align*} But there after if I need to make use of part (1) then I have to interchange the limits. So is it possible. If so, I would like to know what are the conditions we need to have for such an interchange.
Moreover I would like a feedback on the hint given. (Why is it given?)

Answer 1

Okay, so please let me state an analysis principle that will be very useful in numerous problems.

Interversion is painful.

Either there is a straightforward (« logically tautological », that is, that does not require any analysis) argument for doing it, or it requires thinking the whole reasoning through with more elaborate arguments (such as some form of uniform convergence).

So now, let $M$ be a bound for $f$. Let $\epsilon >0$. I suggest that you prove that

$$\left|\int_0^1{f(x^n)}-f(0)\right| \leq \left|\int_0^{1-\epsilon}{f(x^n)}-(1-\epsilon)f(0)\right| + M\epsilon + \epsilon |f(0)|. $$

Thus for every $n$ large enough $$\left|\int_0^1{f(x^n)}-f(0)\right| \leq 3M\epsilon$$.

Answer 2

Another approach uses the mean value theorem for integrals. Indeed, set $g_n(x)=f(x^n).$ Then, there is a sequence $(c_n)\subseteq (0,1-\epsilon)$ such that $\int\limits_0^{1-\epsilon}g_n(x)dx=g_n(c_n)(1-\epsilon)=f(c^n_n)(1-\epsilon).$ Then, $c_n^n\to 0$ as $n\to \infty$ and so $f(c_n^n)\to f(0).$

For the second part, note that $\int_0^1f(x^n)\,dx=\int_0^{1-\epsilon}f(x^n)\,dx+\int_{1-\epsilon}^1f(x^n)\,dx,$ so we just need to consider the second integral. But as $g_n$ is bounded, we can apply the dominated convergence theorem to write $\lim \int_{1-\epsilon}^1f(x^n)\,dx=\int_{1-\epsilon}^1f(0)dx=\epsilon f(0)$.

Therefore, $\lim \int_0^1f(x^n)\,dx= f(0).$

Answer 3

So you managed to show that for each $\varepsilon\in(0,1)$ you get: $$ \lim\limits_{n\to\infty}\int\limits_0^{1-\varepsilon}f(x^n)\,{\rm d}x=(1-\varepsilon)f(0) $$ So for each $\varepsilon\in(0,1)$ you can now look at the following: $$ \int\limits_0^1f(x^n)\,{\rm d}x=\int\limits_0^{1-\varepsilon}f(x^n)\,{\rm d}x+\int\limits_{1-\varepsilon}^1f(x^n)\,{\rm d}x$$ Since you know that $f$ is bounded (a continuous function on a closed interval) you know there exists some $M>0$ such that: $$ \forall x\in(0,1),-M\leq f(x)\leq M $$ So you can show that for second interval you get: $$ -M\varepsilon\leq\int\limits_{1-\varepsilon}^1f(x^n){\rm d}x\leq M\varepsilon $$ Or in other words: $$ \int\limits_0^{1-\varepsilon}f(x^n)\,{\rm d}x-M\varepsilon\leq \int\limits_0^1f(x^n)\,{\rm d}x\leq \int\limits_0^{1-\varepsilon}f(x^n){\rm d}x+M\varepsilon $$ Taking the limits on all sides as $n\to\infty$ you get using the first part: $$(1-\varepsilon)f(0)-M\varepsilon\leq\lim\limits_{n\to\infty}\int\limits_0^1f(x^n)\,{\rm d}x\leq(1-\varepsilon)f(0)+M\varepsilon $$ Lastly, taking the limit $\varepsilon\to 0$ gives you the result: $$ \lim\limits_{n\to\infty}\int\limits_0^1f(x^n)\,{\rm d}x=f(0) $$

Answer 4

As $f$ is continuous at $0$, given $c>0$, there exists $\delta>0$ such that $|f(x)-f(0)|<c$ whenever $|x|<\delta$. So choose $n_0$ such that $(1-\epsilon)^{n_0}<\delta$. Then, for any $x<1-\epsilon$ and $n\geq n_0$, we have $x^n<\delta$; then $|f(x^n)-f(0)|<c$. Then \begin{align} (1-\epsilon)(f(0)-c)&=\int_0^{(1-\epsilon)}(f(0)-c)\,dx\leq\int_0^{1-\epsilon}f(x^n)\,dx\leq \int_0^{(1-\epsilon)}(f(0)+c)\,dx\\ \ \\ &=(1-\epsilon)(f(0)+c). \end{align} As we can do this for all $c>0$, we get $$ \lim_{n\to\infty} \int_0^{1-\epsilon}f(x^n)\,dx=(1-\epsilon)f(0) $$

For the second part, because $f$ is continuous and $[0,1]$ is compact, $f$ is bounded, say $a\leq f(x)\leq b$ for all $x$. Then, for any $\epsilon>0$, $$ \int_0^1f(x^n)\,dx=\int_0^{(1-\epsilon)}f(x^n)\,dx+\int_{(1-\epsilon)}^1f(x^n)\,dx. $$ For the second integral, $$ \epsilon\, a\leq \int_{(1-\epsilon)}^1f(x^n)\,dx\leq \epsilon\, b. $$ Then $$ \limsup_n\int_0^1f(x^n)\,dx= (1-\epsilon)f(0)+\int_{(1-\epsilon)}^1f(x^n)\,dx\leq(1-\epsilon)f(0)+\epsilon\,b. $$ As this works for all $\epsilon>0$, $$\tag1 \limsup_n\int_0^1f(x^n)\,dx\leq f(0). $$ Similarly, $$ \liminf_n\int_0^1f(x^n)\,dx= (1-\epsilon)f(0)+\int_{(1-\epsilon)}^1f(x^n)\,dx\geq(1-\epsilon)f(0)+\epsilon\,a. $$ As this works for all $\epsilon>0$, $$\tag2 \liminf_n\int_0^1f(x^n)\,dx\geq f(0). $$ Now $(1)$ and $(2)$ together imply that that $\lim_n\int_0^1f(x^n)\,dx$ exists and that $$ \lim_n\int_0^1f(x^n)\,dx=f(0). $$

Answer 5

Another approach would be to use some measure theory:

Clearly for all $x \in [0,1-\epsilon)$ we have that $x^n \to 0$ as $n \to \infty$, and hence $f(x^n) \to f(0)$ by continuity for all such $n$. Now since the Riemann integral exists, it is equal to the Lebesgue integral;

$lim_{n \to \infty}\int_{0}^{1-\epsilon}f(x^n)dx = lim_{n \to \infty}\int_{[0,1-\epsilon]}f(x^n)d\lambda(x) =_{(1)} \int_{[0,1-\epsilon]}f(0)d\lambda(x) = (1-\epsilon)f(0)$, since $f$ is dominated by the constant $sup_{x \in [0,1]}\{|f(x)|\}<\infty$ and thus the dominated convergence Theorem implies (1). The same argument works for the second part of your question, since $f(x^n) \to f(0)$ for all $x \in [0,1)$ and $\{1\}$ is a Lebesgue null set, thus does not change the value of our integral; that yields $lim_{n \to \infty}\int_{0}^{1}f(x^n)dx = f(0)$