# Find $\lim \limits_{n \to \infty} \int_0^1 f(x) g(x^n) dx$

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?

• What does $D^1$ stand for? Commented Jul 10, 2018 at 0:38
• Differentiable once Commented Jul 10, 2018 at 0:41
• 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. Commented Jul 10, 2018 at 0:45
• Then, what about f(x) ? Commented Jul 10, 2018 at 0:54
• 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. Commented Jul 10, 2018 at 2:54

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 $$|I_2| \leq \delta ||f||_{\infty} ||g||_\infty. \tag{1}$$

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 $$|g(x^n) -g(0)| \leq \varepsilon, \ \ \forall x\in[0,1-\delta] \text{ and } \forall n>N,$$ which follows by continuity of $g$ at $0$.

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

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

• thx a miilion, what do you think of the solution of mengdie1982 below? Commented Jul 10, 2018 at 14:50
• 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.
– Hayk
Commented Jul 10, 2018 at 15:01
• Does this mean Lebesgue's theorem is valid only for Lebesgue's integrals? Commented Jul 10, 2018 at 15:36
• @Wolfdale, what theorem are you referring to ?
– Hayk
Commented Jul 10, 2018 at 15:42
• Lebesgue's dominated convergence theorem Commented Jul 10, 2018 at 15:48

$f(x)g(x^{n}) \to f(x) g(0)$ almost everywhere and it is uniformly bounded, so teh limit is $g(0)\int_0^{1} f(x) \, dx$ by Bounded Convergence Theorem.

# 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 $$\lim_{n \to \infty}\int_0^1 \phi_n(x){\rm d}x=\int_0^1 f(x)g(0){\rm d}x=g(0)\int_0^1f(x){\rm d}x.$$

• thx a miilion, what do you think of the solution of Hayk above? Commented Jul 10, 2018 at 14:51