If , f : [ 0 , b] →^(continuous) R , g : R →_(b−periodic) ^(continuous) R ⇒ lim_(n→∞) ∫_0 ^( b) f(x)g(nx)dx=^? (1/b) ∫_0 ^( b) f(x)dx .∫


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Question Number 204372 by mnjuly1970 last updated on 14/Feb/24

$I f, f : [0, b] \overset{c o n t i n u o u s}{\to} R$ $, g : R \underset{b - p e r i o d i c}{\overset{c o n t i n u o u s}{\to}} R$ $\Rightarrow {l i m}_{n \to \infty} \int_{0}^{b} f (x) g (n x) d x \overset{?}{=} \frac{1}{b} \int_{0}^{b} f (x) d x . \int_{0}^{b} g (x) d x$
	Answered by witcher3 last updated on 15/Feb/24
	$Ω=lim_(n→∞) ∫_0 ^b f(x)g(nx)dx=(1/b)∫_0 ^b f(x)dx.∫_0 ^b g(x)dx nx=y⇒Ω=lim_(n→∞) (1/n)∫_0 ^(nb) f((y/n))g(y)dy=lim_(n→∞) (1/n)Σ_(k=1) ^(n−1) ∫_(kb) ^((k+1)b) f((y/n))g(y)dy y=kb+z;g(y)=g(z) by b periodicity of g =lim_(n→∞) (1/n)Σ_(k=1) ^(n−1) ∫_0 ^b f(((kb)/n)+(z/n))g(z)dz =∫_0 ^b lim_(n→∞) (1/n)Σ_(k=1) ^(n−1) f(((kb)/n)+(z/n))g(z)dz let f^∼ (x)=f(x+(z/n)) f^∼ (x) cv uniformly to f lim_(n→∞) f(x+(z/n))=f(x) Ω=∫_0 ^b lim_(n→∞) (1/n)Σ_(k=0) ^(n−1) f^∼ (((k(b−0))/n)).g(z)dz=(1/b)∫_0 ^b lim_(n→∞) ((b−0)/n)∫f(((kb)/n))g(z)dz =(1/b).lim_(n→∞) {(b/n)f(((kb)/n))}.∫_0 ^b g(z)dz Σ_(k=0) ^(n−1) ((b−0)/n)f(k(b/n))=∫_0 ^b f(x)dx Ω=(1/b)∫_0 ^b f(x)dx.∫_0 ^b g(z)dz,z muet variable =(1/b)∫_0 ^b f(x)dx.∫_0 ^b g(x)dx riemann cv by contonuity of f lim_(n→∞) ∫_0 ^b f^∼ (x)dx=∫_0 ^b f(x)dx by uniforme cv of f^∼ (x) lim_(n→∞) sup∣f(x+(z/n))−f(x)∣=0 “since f is defined [0,b] compact⇒simple cv⇒uniforme cv”$
	$Ω = \lim_{n \to \infty} \int_{0}^{b} f (x) g (nx) dx = \frac{1}{b} \int_{0}^{b} f (x) dx . \int_{0}^{b} g (x) dx$ $nx = y \Rightarrow Ω = \lim_{n \to \infty} \frac{1}{n} \int_{0}^{nb} f (\frac{y}{n}) g (y) dy = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n - 1}} \int_{kb}^{(k + 1) b} f (\frac{y}{n}) g (y) dy$ $y = kb + z; g (y) = g (z) by b periodicity of g$ $= \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n - 1}} \int_{0}^{b} f (\frac{kb}{n} + \frac{z}{n}) g (z) dz$ $= \int_{0}^{b} \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n - 1}} f (\frac{kb}{n} + \frac{z}{n}) g (z) dz$ $let \tilde{f} (x) = f (x + \frac{z}{n})$ $\tilde{f} (x) cv uniformly to f$ $\underset{n \to \infty}{\lim f} (x + \frac{z}{n}) = f (x)$ $Ω = \int_{0}^{b} \lim_{n \to \infty} \frac{1}{n} \underset{k = 0}{\sum^{n - 1}} \tilde{f} (\frac{k (b - 0)}{n}) . g (z) dz = \frac{1}{b} \int_{0}^{b} \lim_{n \to \infty} \frac{b - 0}{n} \int f (\frac{kb}{n}) g (z) dz$ $= \frac{1}{b} . \lim_{n \to \infty} {\frac{b}{n} f (\frac{kb}{n})} . \int_{0}^{b} g (z) dz$ $\underset{k = 0}{\sum^{n - 1}} \frac{b - 0}{n} f (k \frac{b}{n}) = \int_{0}^{b} f (x) dx$ $Ω = \frac{1}{b} \int_{0}^{b} f (x) dx . \int_{0}^{b} g (z) dz, z muet variable$ $= \frac{1}{b} \int_{0}^{b} f (x) dx . \int_{0}^{b} g (x) dx$ $riemann cv by contonuity of f$ $\lim_{n \to \infty} \int_{0}^{b} \tilde{f} (x) dx = \int_{0}^{b} f (x) dx$ $by uniforme cv of \tilde{f} (x)$ $\lim_{n \to \infty} \sup ∣ f (x + \frac{z}{n}) - f (x) ∣= 0$ $‘ ‘ since f is defined [0, b] compact \Rightarrow simple cv \Rightarrow uniforme {cv}^{″}$
	Commented by mnjuly1970 last updated on 15/Feb/24

	$t h a n k y o u s o m u c h$ $s i r w i c h e r . e x c e l l e n t p r o o f .$
	Answered by witcher3 last updated on 15/Feb/24

	$nice probleme didnt expcte such result existe$
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