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For positive integer $n$, let $f_n$ be the function defined on the interval $[0,1]$ by $f_n(x) = \frac{x^n}{1+x^n}$. Which of the following statements are true?

  • The sequence $\{f_n\}$ converges pointwise on $[0,1]$ to a limit function $f$.
  • The sequence $\{f_n\}$ converges uniformly on $[0,1]$ to a limit function $f$.
  • $$\lim_{n \to \infty}\int_0^1f_n(x)dx = \int_0^1 \left(\lim_{n\to\infty}f_n(x)\right)dx$$

I verified that $f_n$ is pointwise convergence but not uniformly convergence, then I assumed that the exchange is only valid for uniformly convergence and marked the third option as false. However, the answer indicates that the third option is true. So what exactly is the condition for the exchange of limit and integral?

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  • $\begingroup$ Dominated convergence Theorem with dominating function $1$. Or Monotone Convergence Theorem applied to $1-f_n$. $\endgroup$ Commented Oct 17, 2023 at 12:03
  • $\begingroup$ @geetha290krm I see, so I guess I shouldn't have used convergence uniformly or pointwise to conclude the result in the first place. $\endgroup$
    – OriginK
    Commented Oct 17, 2023 at 12:06
  • $\begingroup$ For a proof without Measure Theory, use uniform convergence on $[0,r]$ for $r<1$ and see what happens to the integrals from $r $ to $1$ when $r$ is close to $1$. $\endgroup$ Commented Oct 17, 2023 at 12:07
  • $\begingroup$ @OriginK uniform convergence is sufficient to switch the limit and integral but not necessary. $\endgroup$ Commented Oct 17, 2023 at 12:13
  • $\begingroup$ @geetha290krm I see. It seems that only the end point is the problem. So is it true that given a pointwise convergence like this, I can claim the exchange of the limit and the integral sign as long as one point is the problem (and maybe it doesn't blow up?) $\endgroup$
    – OriginK
    Commented Oct 17, 2023 at 12:22

2 Answers 2

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An alternative method to the one presented in the previous answer is the following: $$0<\int_0^1 \frac{x^n}{1+x^n}dx<\int_0^1x^ndx=\frac{1}{n+1}$$ Hence, by the squeeze theorem, you conclude that $$\lim_{n\to +\infty}\int_0^1\frac{x^n}{1+x^n}dx=0$$

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We can show that the limit can be interchanged by just computing the two (I'm assuming you don't know the dominated convergence theorem). Indeed as you probably noticed,

$$\lim_{n\to\infty}f_n(x)=\begin{cases} 0,&x\in[0,1),\\ \frac{1}{2},&x=1. \end{cases}$$

Consequently

$$\int_0^1\lim_{n\to\infty}f_n(x)~\mathrm{d}x=0.$$

We thus have to show that

$$\lim_{n\to\infty}\int_0^1 f_n(x)~\mathrm{d}x=0.$$

Notice that, for any $\varepsilon\in(0,1)$ and all $x\in[0,\varepsilon]$

$$0\leq f_n(x)\leq\frac{\varepsilon^n}{1+\varepsilon^n},$$

and for any $x\in(\varepsilon,1]$,

$$0\leq f_n(x)\leq 1.$$

Consequently

$$0\leq\int_0^1f_n(x)~\mathrm{d}x=\int_0^\varepsilon f_n(x)~\mathrm{d}x+\int_\varepsilon^1 f_n(x)~\mathrm{d}x\leq \frac{\varepsilon^{n+1}}{1+\varepsilon^n}+(1-\varepsilon)\to(1-\varepsilon)$$

as $n\to\infty$. But as $\varepsilon\in(0,1)$ was arbitrary, it follows that

$$\lim_{n\to\infty}\int_0^1f_n(x)~\mathrm{d}x=0.$$

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  • $\begingroup$ Thank you for your detailed explanation! $\endgroup$
    – OriginK
    Commented Oct 17, 2023 at 12:19

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