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Find $$\lim \limits_{n \to \infty} \int_0^1 f(x) g(x^n) dx$$ Where $\;f\in C^0, \quad g\in D^1$

Any hints?

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  • $\begingroup$ What does $D^1$ stand for? $\endgroup$ Jul 10, 2018 at 0:38
  • $\begingroup$ Differentiable once $\endgroup$
    – Wolfdale
    Jul 10, 2018 at 0:41
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    $\begingroup$ Notice that $x^n$ converges to $0$ for all $x\in[0,1)$. Can you utilize this information for your problem? Guessing the answer would be straightforward, while establishing convergence rigorously may be not so immediate, depending on your background. $\endgroup$ Jul 10, 2018 at 0:45
  • $\begingroup$ Then, what about f(x) ? $\endgroup$
    – Wolfdale
    Jul 10, 2018 at 0:54
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    $\begingroup$ Is $g$ differentiable once, with continuous derivative? Note that this allows you to apply integration by parts to the product of $f$ and $g(x^n)$, since $g(x^n)$ is also a once differentiable function. Now deal with the two parts separately. $\endgroup$ Jul 10, 2018 at 2:54

3 Answers 3

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You don't really need $g$ to be differentiable, it is enough to assume $g \in C[0,1]$ (in fact, all you need is to have $g$ integrable and bounded, and continuous at $0$, as the proof below shows). Then the limit in question equals $$ g(0) \int_0^1 f(x) dx, $$ which we prove below.

Fix $\delta>0$ small and split the integral into $2$ parts, namely $$ \int_0^1 f(x) g(x^n) dx = \int_0^{1-\delta} + \int_{1-\delta}^1 := I_1 + I_2. $$

In view of continuity of $f$ and $g$, both functions are bounded on $[0,1]$ and we get \begin{equation} |I_2| \leq \delta ||f||_{\infty} ||g||_\infty. \tag{1} \end{equation}

For $I_1$, observe that $x\in [0,1-\delta]$, hence for any $\varepsilon>0$ small by choosing $N\in \mathbb{N}$ large enough we get \begin{equation} |g(x^n) -g(0)| \leq \varepsilon, \ \ \forall x\in[0,1-\delta] \text{ and } \forall n>N, \end{equation} which follows by continuity of $g$ at $0$.

From here, for all $n>N$ we obtain \begin{equation} \left| I_1 - g(0) \int_{0}^{1-\delta} f(x) dx \right| \leq \int_0^{1-\delta} |f(x)| |g(x^n) - g(0)| dx \leq \varepsilon \int_0^1|f(x)| dx. \tag{2} \end{equation}

Since $\delta>0$ and $\varepsilon>0$ are arbitrary small numbers, from $(1)$ and $(2)$ we get the claim.

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  • $\begingroup$ thx a miilion, what do you think of the solution of mengdie1982 below? $\endgroup$
    – Wolfdale
    Jul 10, 2018 at 14:50
  • $\begingroup$ you're welcome @Wolfdale; as for your question, I think one can get along without a recourse to dominated convergence theorem (or alike), to get a self-contained treatment for the problem; this is not an opinion, but a mathematical comment. $\endgroup$
    – Hayk
    Jul 10, 2018 at 15:01
  • $\begingroup$ Does this mean Lebesgue's theorem is valid only for Lebesgue's integrals? $\endgroup$
    – Wolfdale
    Jul 10, 2018 at 15:36
  • $\begingroup$ @Wolfdale, what theorem are you referring to ? $\endgroup$
    – Hayk
    Jul 10, 2018 at 15:42
  • $\begingroup$ Lebesgue's dominated convergence theorem $\endgroup$
    – Wolfdale
    Jul 10, 2018 at 15:48
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$f(x)g(x^{n}) \to f(x) g(0)$ almost everywhere and it is uniformly bounded, so teh limit is $g(0)\int_0^{1} f(x) \, dx$ by Bounded Convergence Theorem.

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Solution

We only need to assume that:

  • $f$ is integrable over $[0,1]$;
  • $g$ is bounded over $[0,1]$ and continuous at $x=0$.

Thus, denote $\phi_n(x)=f(x)g(x^n)$ where $x \in [0,1],n=1,2,\cdots$, and $\max\limits_{0\leq x \leq 1} |g(x)|=M$. It's easy to have that

$$|\phi_n(x)| \leq M\cdot|f(x)|,$$ for $n=1,2,\cdots.$ Moreover, the dominating function $M \cdot|f(x)|$ is obviously integrale over $[0,1].$ Besides, $$\lim_{n \to \infty}\phi_n(x)=f(x)g(0),~~~~x \in [0,1].$$

As a result, by Lebesgue's dominated convergence theorem, we have $$\lim_{n \to \infty}\int_0^1 \phi_n(x){\rm d}x=\int_0^1 f(x)g(0){\rm d}x=g(0)\int_0^1f(x){\rm d}x.$$

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  • $\begingroup$ thx a miilion, what do you think of the solution of Hayk above? $\endgroup$
    – Wolfdale
    Jul 10, 2018 at 14:51

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