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)
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)
Commented by mnjuly1970 last updated on 15/Feb/24
Answered by witcher3 last updated on 15/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-