Question Number 204372 by mnjuly1970 last updated on 14/Feb/24
![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 .∫_0 ^( b) g(x)dx](https://www.tinkutara.com/question/Q204372.png)
$$ \\ $$$$\:\:{If}\:,\:\:\:\:{f}\::\:\left[\:\mathrm{0}\:,\:{b}\right]\:\overset{{continuous}} {\rightarrow}\:\mathbb{R}\: \\ $$$$\:\:\:\:\:\:\:\:,\:\:\:\:{g}\::\:\mathbb{R}\:\underset{{b}−{periodic}} {\overset{{continuous}} {\rightarrow}}\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:{lim}_{{n}\rightarrow\infty} \:\int_{\mathrm{0}} ^{\:{b}} {f}\left({x}\right){g}\left({nx}\right){dx}\overset{?} {=}\frac{\mathrm{1}}{{b}}\:\int_{\mathrm{0}} ^{\:{b}} {f}\left({x}\right){dx}\:.\int_{\mathrm{0}} ^{\:{b}} {g}\left({x}\right){dx} \\ $$$$ \\ $$
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”](https://www.tinkutara.com/question/Q204385.png)
$$\Omega=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{nx}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{b}}\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}.\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{g}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{nx}=\mathrm{y}\Rightarrow\Omega=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\int_{\mathrm{0}} ^{\mathrm{nb}} \mathrm{f}\left(\frac{\mathrm{y}}{\mathrm{n}}\right)\mathrm{g}\left(\mathrm{y}\right)\mathrm{dy}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\int_{\mathrm{kb}} ^{\left(\mathrm{k}+\mathrm{1}\right)\mathrm{b}} \mathrm{f}\left(\frac{\mathrm{y}}{\mathrm{n}}\right)\mathrm{g}\left(\mathrm{y}\right)\mathrm{dy} \\ $$$$\mathrm{y}=\mathrm{kb}+\mathrm{z};\mathrm{g}\left(\mathrm{y}\right)=\mathrm{g}\left(\mathrm{z}\right)\:\:\mathrm{by}\:\mathrm{b}\:\mathrm{periodicity}\:\mathrm{of}\:\mathrm{g} \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\frac{\mathrm{kb}}{\mathrm{n}}+\frac{\mathrm{z}}{\mathrm{n}}\right)\mathrm{g}\left(\mathrm{z}\right)\mathrm{dz} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{b}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{f}\left(\frac{\mathrm{kb}}{\mathrm{n}}+\frac{\mathrm{z}}{\mathrm{n}}\right)\mathrm{g}\left(\mathrm{z}\right)\mathrm{dz} \\ $$$$\mathrm{let}\:\overset{\sim} {\mathrm{f}}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}+\frac{\mathrm{z}}{\mathrm{n}}\right) \\ $$$$\overset{\sim} {\mathrm{f}}\left(\mathrm{x}\right)\:\mathrm{cv}\:\mathrm{uniformly}\:\mathrm{to}\:\mathrm{f} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}f}\left(\mathrm{x}+\frac{\mathrm{z}}{\mathrm{n}}\right)=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{b}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\overset{\sim} {\mathrm{f}}\left(\frac{\mathrm{k}\left(\mathrm{b}−\mathrm{0}\right)}{\mathrm{n}}\right).\mathrm{g}\left(\mathrm{z}\right)\mathrm{dz}=\frac{\mathrm{1}}{\mathrm{b}}\int_{\mathrm{0}} ^{\mathrm{b}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{b}−\mathrm{0}}{\mathrm{n}}\int\mathrm{f}\left(\frac{\mathrm{kb}}{\mathrm{n}}\right)\mathrm{g}\left(\mathrm{z}\right)\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{b}}.\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{b}}{\mathrm{n}}\mathrm{f}\left(\frac{\mathrm{kb}}{\mathrm{n}}\right)\right\}.\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{g}\left(\mathrm{z}\right)\mathrm{dz} \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{b}−\mathrm{0}}{\mathrm{n}}\mathrm{f}\left(\mathrm{k}\frac{\mathrm{b}}{\mathrm{n}}\right)=\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{b}}\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}.\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{g}\left(\mathrm{z}\right)\mathrm{dz},\mathrm{z}\:\mathrm{muet}\:\mathrm{variable} \\ $$$$=\frac{\mathrm{1}}{\mathrm{b}}\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}.\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{g}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{riemann}\:\mathrm{cv}\:\mathrm{by}\:\mathrm{contonuity}\:\mathrm{of}\:\mathrm{f} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{b}} \overset{\sim} {\mathrm{f}}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{by}\:\mathrm{uniforme}\:\mathrm{cv}\:\mathrm{of}\:\overset{\sim} {\mathrm{f}}\left(\mathrm{x}\right) \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sup}\mid\mathrm{f}\left(\mathrm{x}+\frac{\mathrm{z}}{\mathrm{n}}\right)−\mathrm{f}\left(\mathrm{x}\right)\mid=\mathrm{0} \\ $$$$“\mathrm{since}\:\mathrm{f}\:\mathrm{is}\:\mathrm{defined}\:\left[\mathrm{0},\mathrm{b}\right]\:\mathrm{compact}\Rightarrow\mathrm{simple}\:\mathrm{cv}\Rightarrow\mathrm{uniforme}\:\mathrm{cv}'' \\ $$
Commented by mnjuly1970 last updated on 15/Feb/24
$${thank}\:{you}\:{so}\:{much}\: \\ $$$${sir}\:{wicher}\:.{excellent}\:{proof}. \\ $$
Answered by witcher3 last updated on 15/Feb/24
$$\mathrm{nice}\:\mathrm{probleme}\:\mathrm{didnt}\:\mathrm{expcte}\:\mathrm{such}\:\mathrm{result}\:\mathrm{existe} \\ $$