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Let $f(x)$ be continuous on the closed interval $[0,1]$ and satisfy $f(0)=1$, $f(1/2)=2$, $f(1)=3$. Show that $$\lim_{n \to \infty} \int_0^1 f(x^n) \ dx$$ exists and compute its limit.

I plotted the function $2x^n+1$ on desmos and I see that most values assume $1$ as $n \to \infty$. So, my guess is that the integral is going to approach $1$.

If we also compute $\int_0^1 (2x^n+1) \ dx=\frac{2x^{n+1}}{n+1}+x\Big|_0^1=\frac{2}{n+1}+1$, we see that to $\frac{2}{n+1}+1 \to 1$ as $n \to \infty$.

How can I make this rigorous?

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For each $x \in (0,1)$, $x^n \to 0$ as $n \to \infty$, so by continuity, $f(x^n) \to f(0)$. Since $f$ is continuous on $[0,1]$ it is bounded, say by some constant $C$, so that $\forall x,n$, $|f(x^n)| \le C$. Since $C$ is integrable on $[0,1]$, dominated convergence theorem tells us $\int_0^1 f(x^n)dx \to \int_0^1 f(0)dx = 1$.

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By change of variable we have \begin{align*} \int_{0}^{1}f(x^{n})dx=\dfrac{1}{n}\int_{0}^{1} u^{n^{-1}-1}f(u)du, \end{align*} and we also know that \begin{align*} \int_{0}^{1}u^{n^{-1}-1}du=n, \end{align*} so \begin{align*} \dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}f(u)du-1&=\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}f(u)du-\dfrac{1}{n}\int_{0}^{1}f(0)u^{n^{-1}-1}du\\ &=\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}(f(u)-f(0))du. \end{align*} Given $\epsilon>0$, since $f$ is continuous, there exists some $\delta\in(0,1)$ such that for all $u\in(0,\delta)$, we have $|f(u)-f(0)|<\epsilon$. Now, \begin{align*} &\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}|f(u)-f(0)|du\\ &=\dfrac{1}{n}\int_{0}^{\delta}u^{n^{-1}-1}|f(u)-f(0)|du+\dfrac{1}{n}\int_{\delta}^{1}u^{n^{-1}-1}|f(u)-f(0)|du\\ &\leq\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}\cdot\epsilon du+\dfrac{2M}{n}\int_{\delta}^{1}u^{n^{-1}-1}du\\ &=\epsilon+\dfrac{2M}{n}\cdot n(1-\delta^{n^{-1}})\\ &=\epsilon+2M(1-\delta^{n^{-1}}), \end{align*} where $M:=\sup_{x\in[0,1]}|f(x)|$. We have $\delta^{n^{-1}}\rightarrow 1$ as $n\rightarrow\infty$, so \begin{align*} \dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}|f(u)-f(0)|du\rightarrow 0, \end{align*} and hence \begin{align*} \dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}f(u)du\rightarrow 1. \end{align*}

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  • $\begingroup$ You used uniform continuity to get continuity at 0.... $\endgroup$ – mathworker21 Dec 16 '17 at 3:59
  • $\begingroup$ That's simply redundant, I have changed that. $\endgroup$ – user284331 Dec 16 '17 at 4:00
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Since $f$ is continuous on a compact set its bounded by some constant function $M$. Infact, choosing the sequence of measurable functions $ f_n(x):= f(x^n)$ we have

$$ |f_n(x)| \leq M \ \forall x , n $$

Also notice $f_n$ converges to $1$ pointwise for $x \in [0,1)$.

By the dominated convergence theorem

$$ \lim_{n \rightarrow \infty} \int f_n = \int 1 = 1$$

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  • $\begingroup$ I was thinking about choosing $f_n(x)=f(x^n)$. Is the dominated convergence theorem the theorem that tells us that limit and integral commute? $\endgroup$ – Al Jebr Dec 16 '17 at 3:22
  • $\begingroup$ I just saw dominated convergence theorem is used in Lebesgue integration. I haven't covered this yet. $\endgroup$ – Al Jebr Dec 16 '17 at 3:23
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Without the dominated convergence theorem:

  • First, $f$ is continuous, so is bounded on $[0,1]$: Let$C>0$ be such that $\sup_{[0,1]} \lvert f\rvert + 1 \leq C$. Also, $x\mapsto f(x^n)$ is continuous as well, so the integral $\int_0^1 f(x^n)dx$ is well-defined for all $n\in\mathbb{N}$.

  • Fix $\varepsilon > 0$, and let $\delta$ be such that (by continuity at $0$) $$\lvert f(x) - 1\rvert \leq \varepsilon$$ for all $x\in[0,1]$ such that $\lvert x\rvert \leq \delta$.

  • Consider $n_0\in\mathbb{N}$ such that $\left(1-\frac{\varepsilon}{C}\right)^n \leq \delta$ for all $n\geq n_0$.

Then, for all $n\geq n_0$, we have $\lvert x^n\rvert \leq \delta$ for all $x\in[0,1-\frac{\varepsilon}{C}]$, and we will use that to handle $$ \int_0^1 f(x^n)dx = \int_0^{1-\frac{\varepsilon}{C}} f(x^n)dx + \int_{1-\frac{\varepsilon}{2C}}^1 f(x^n)dx \tag{1} $$ For the second term: $$ \left\lvert \int_{1-\frac{\varepsilon}{2C}}^1 f(x^n)dx\right\rvert \leq \int_{1-\frac{\varepsilon}{2C}}^1 \left\lvert f(x^n)\right\rvert dx \leq \int_{1-\frac{\varepsilon}{2C}}^1 Cdx = \varepsilon\,. \tag{2} $$

For the first term: $$ 1\leq \left(1-\frac{\varepsilon}{C}\right)\left(1+\varepsilon\right) \leq \int_0^{1-\frac{\varepsilon}{C}} \left(1-\varepsilon\right)dx \leq \int_0^{1-\frac{\varepsilon}{C}} f(x^n)dx\leq \int_0^{1-\frac{\varepsilon}{2C}} \left(1+\varepsilon\right)dx \leq 1+\varepsilon. \tag{3} $$ using continuity at $0$ and our choice of $n_0$.

Overall, using (1), (2), and (3), $$ 1-2\varepsilon \leq \int_0^1 f(x^n)dx \leq 1+2\varepsilon $$ for all $n\geq n_0$. This shows the convergence to $1$, as $\varepsilon$ was arbitrary.

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  • $\begingroup$ Why did you choose $1-\frac{\epsilon}{c}$? $\endgroup$ – Al Jebr Jul 3 '18 at 20:08
  • $\begingroup$ @AlJebr In order to establish (2), to handle $C\cdot$(interval of integration). $\endgroup$ – Clement C. Jul 3 '18 at 20:11

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