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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?)

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

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  • $\begingroup$ Thank you @Mindlack.So this means we are proving the limit definition for this one as well. ? Also I think if we choose the interval as $(0,\frac{\epsilon}{3})$ we would end up in epsilon alone in the RHS $\endgroup$ – DD90 Jan 6 at 23:14
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    $\begingroup$ Yes, unless you can use some more advanced machinery (double-limit theorems, uniform convergence and stuff), you have to prove the limit using $\epsilon$. On another note: if you wanted to have some $\epsilon$ in the final estimate, you would have to take the integral to $1-\frac{\epsilon}{3M+1}$. $\endgroup$ – Mindlack Jan 6 at 23:18
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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 $$ (1-\epsilon)(f(0)-c)=\int_0^{(1-\epsilon)}(f(0)-c)\,dx\leq\int_0^{1-\epsilon}f(x^n)\,dx\leq \int_0^{(1-\epsilon)}(f(0)+c)\,dx=(1-\epsilon)(f(0)+c). $$ As we can do this for all $c>0$, we get $$ \int_0^{1-\epsilon}f(x^n)\,dx=(1-\epsilon)f(0) $$

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

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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: $$ \lim\limits_{n\to\infty}\int\limits_0^1f(x^n)\,{\rm d}x=f(0) $$

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

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  • $\begingroup$ why the downvote? since $(c_n)\subseteq (0,1-\epsilon),\ \limsup c_n^n\to 0$ and since $f$ is continuous, $f(c_n^n)\to 0$. $\endgroup$ – Matematleta Jan 6 at 23:16

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