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Let $f: [0,1] \to \mathbb{R}$ be continuous, prove $\lim_{n\to \infty} \int_0^1 f(x^n)dx = f(0)$

This makes some sense looking at it. I have only the Regulated Integral definition of integration to work with :https://en.wikipedia.org/wiki/Regulated_integral

It uses sequences of step functions with a partition over a closed interval.

So I need to solve the integral prior to taking the limit but not sure how to really get what I need. Since $f $ is continuous function there exists a sequence $(\varphi_n)_{n \in \mathbb{N}}$ of step functions such that $\lim_{n \to \infty} \sup_{x \in [0,1]} \mid f(x) - \varphi_n(x)\mid \,= 0$. Moreover, $\int_0^1 f(x)dx := \lim_{m \to \infty} \int_0^1 \varphi_m(x)dx $ so it should be the case that $\int_0^1 f(x^n)dx := \lim_{m \to \infty} \int_0^1 \varphi_m(x^n)dx $ so I assume that it would be that $\lim_{n \to \infty } \int_0^1 f(x^n)dx := \lim_{n \to\infty} (\lim_{n \to \infty} \int_0^1 \varphi_m(x^n)dx )$

Now, $ \int_a^b \varphi(\eta)d\eta := \sum_{j=0}^{N}\varphi(\eta_j)(\sigma_{j+1}-\sigma_j)$ where $(\sigma_j)_{j=0}^{N+1})$ is a partition of $[a,b]$ and $\eta_{j} \in (\sigma_j,\sigma_{j+1})$ such that each block of the partition is constant so choice of $\eta_j$ is not particularly important.

So should have $ lim_{n \to \infty}(lim_{m \to \infty}\int_a^b \varphi_m(\eta^n)d\eta := lim_{n \to \infty}(\lim_{m \to \infty}\sum_{j=0}^{N}\varphi_m(\eta_j^n)(\sigma_{j+1}-\sigma_j))$.

This is where I get stuck.

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    Just to check: you are *not* familiar with the Dominated Convergence Theorem, I assume? (This would give a rather concise proof, even if it is overkill). – 2017-01-17
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    @ClementC. Man, I just love overkill! Kill the cockroaches with nukes by all means!!! – 2017-01-17
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    @ClementC. No I am not, I will look it up but I can't use it if we haven't covered it yet. – 2017-01-17
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    @ClementC. Ah yeah, I am not ready for that – 2017-01-17
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    @oliverjones Without summoning the full artillery, you can also give a more basic $\varepsilon$-$\delta$ based argument -- see my answer. – 2017-01-17
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    Clement provided an excellent answer, but here's perhaps a less confusing argument: suppose WLOG that $f(0) = 0$. Using the change of variables $u=x^n$, we get: $$\frac1n \int_0^1 \frac{f(x)}{x^{1-\frac1n}} dx:=I(n)$$ Let $\epsilon >0$ and choose $\delta < 1$ such that $|f(x)|<\epsilon/2$ for $0\le x < \delta$. The quantity $\int_{\delta}^1 \frac{f(x)}{x^{1-\frac1n}} dx$ is absolutely bounded by a constant, so choose $N$ such that for $n\ge N$, $\frac1n \int_{\delta}^1 \frac{f(x)}{x^{1-\frac1n}} dx < \epsilon/2$. It follows that for $n\ge N$, $|I(n)| <\epsilon$. – 2017-01-17

2 Answers 2

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As I mentioned in a comment, if you allow the Dominated Convergence Theorem then this is a matter of a few lines. The sequence of functions $(f_n)_n$ defined by $f_n(x) \stackrel{\rm def}{=} f(x^n)$ ($x\in[0,1])$ converges pointwise (by continuity of $f$) to the constant function $f(0)$ on $[0,1)$, and one can uniformly dominate all $f_n$'s by the integrable (because constant) function $g\stackrel{\rm def}{=} \lVert f\rVert_\infty$ ($f$ being bounded on $[0,1]$, since it is continuous). Now, by the DCT we get $$ \int_0^1 f_n \xrightarrow[n\to\infty]{} \int_0^1 \lim_{n\to\infty}f_n = \int_0^1 f(0)dx = f(0). $$

But this is overkill. Another option, explored below, would be to use continuity of $f$ at $0$: recall that $f$ being continuous on a closed interval, it is bounded.

Without loss of generality, we can assume $f(0) = 0$ (indeed, just apply the same exercise to $g\stackrel{\rm def}{=} f-f(0)$. It is easy to check that this will be sufficient to prove the result).

Fix $\varepsilon>0$, break the interval in 2: $$ \int_0^1 f(x^n)dx = \int_0^a f(x^n)dx +\int_a^1 f(x^n)dx $$ for some convenient $a=a(\varepsilon, f) \stackrel{\rm def}{= } 1-\frac{\varepsilon}{2\lVert f\rVert_\infty}$. The second part has absolute value at most $(1-a)\cdot\lVert f\rVert_\infty < \frac{\varepsilon}{2}$ by definition.

As for the first, we use continuity of $f$ at $0$: let $\delta_\varepsilon>0$ be such that $\lvert u\rvert \leq \delta_\varepsilon$ implies $\lvert f(u)\rvert \leq \frac{\varepsilon}{2}$. We then rely on the fact that for $n$ big enough (i.e., for $n\geq N_{\varepsilon}$, for some $N_{\varepsilon}\geq 0$, we have $\lvert x^n\rvert \leq \lvert a^n\rvert \leq \delta_\varepsilon$. Thus, the first term can be bounded as $$ \left\lvert \int_0^a f(x^n)dx\right\rvert \leq \int_0^a \left\lvert f(x^n)\right\rvert dx \leq \int_0^a \frac{\varepsilon}{2} dx \leq \frac{\varepsilon}{2} $$ for any $n\geq N_{\varepsilon}$.

Putting it together, for $n\geq N_{\varepsilon}$ one has $$ \left\lvert \int_0^1 f(x^n)dx\right\rvert \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$ by the triangle inequality. This shows that $$ \int_0^1 f(x^n)dx\xrightarrow[n\to\infty]{} 0 $$ and from there the general case follows.

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Since $f $ is continuous, given $\varepsilon>0$ there exists $\delta>0$ such that $|f (x)-f (0)|<\varepsilon/2 $ whenever $|x|<\delta $. Let $M=\max \{|f (x)|:\ x\in [0,1]\}$. Choose $x_0\in (0,1) $ such that $1-x_0 <\varepsilon/2M $. Choose $n_0$ such that $|x_0^{n_0}|<\delta $ . Then, for any $n\geq n_0$,

\begin{align} \left|\int_0^1f (x^n)\,dx-f (0)\right| &=\left|\int_0^1(f (x^n)-f (0))\,dx\right| \leq \int_0^1|f (x^n)-f (0)|\,dx \\ \ \\ &= \int_0^{x_0}|f (x^n)-f (0)|\,dx + \int_{x_0}^1|f (x^n)-f (0)|\,dx \\ \ \\ &\leq \frac\varepsilon2 x_0+2M (1-x_0)\\ \ \\ &<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon. \end{align}

As an aside, if you have the Fundamental Theorem of Calculus available, the substitution $u=x^n $ does the job.

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    Isn't it exactly the same argument as in my answer? – 2017-01-17
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    very nice proof, professor as you made it much easier to follow as an undergrad student at a state university. – 2017-01-17