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Question Number 2032 by 123456 last updated on 31/Oct/15
f:[0,+1]→R η(f):=∫_0 ^1 f^2 dx and if ∀x∈[0,1],f∈[m,M] for some (m,M)∈R^2 f↑:=max(f)−f f↓:=f−min(f) μ(f):=∫_0 ^1 f↓f↑dx then f=e^x , η(f)=^? μ(f) f=x,η(f)=^? μ(f) ∃f,η(f)=0??? ∃f,μ(f)=0??? μ(f)=0, does η(f)=0??? η(f)=0, does μ(f)=0???
f:[0,+1]→Rη(f):=∫10f2dxandif∀x∈[0,1],f∈[m,M]forsome(m,M)∈R2f↑:=max(f)−ff↓:=f−min(f)μ(f):=∫10f↓f↑dxthenf=ex,η(f)=?μ(f)f=x,η(f)=?μ(f)∃f,η(f)=0???∃f,μ(f)=0???μ(f)=0,doesη(f)=0???η(f)=0,doesμ(f)=0???
Commented by Yozzi last updated on 31/Oct/15
An attempt max(f)=M,min(f)=m ∴ f↑=M−f , f↓=f−m μ(f)=∫_0 ^1 f↑f↓dx=∫_0 ^1 (M−f)(f−m)dx μ(f)=∫_0 ^1 (M+m)fdx−∫_0 ^1 f^2 dx−Mm∫_0 ^1 dx μ(f)=(M+m)∫_0 ^1 fdx−η(f)−Mm μ(f)+η(f)=(M+m)∫_0 ^1 fdx−Mm f(x)=e^x η(e^x )=∫_0 ^1 e^(2x) dx=(1/2)(e^2 −1)=(1/2)(e+1)(e−1) ∀x∈[0,1],e^x ∈[1,e]⇒m=1,M=e ∴μ(e^x )=(1+e)∫_0 ^1 e^x dx−(1/2)(e+1)(e−1)−e μ(e^x )=(1+e)(e−1)−0.5(e+1)(e−1)−e μ(e^x )=0.5(e+1)(e−1)−e≠0.5(e−1)(e+1) μ(e^x )≠η(e^x ) f(x)=x η(x)=∫_0 ^1 x^2 dx=(1/3) ∀x∈[0,1], f∈[0,1]⇒m=0,M=1. ∴μ(x)=(0+1)∫_0 ^1 xdx−(1/3)−0 μ(x)=(1/2)−(1/3)=(1/6)≠(1/3) ∴μ(x)≠η(x) For f(x)≠0 ∀x∈[0,1], if η(f)=0⇒f^2 is odd for x∈[0,1] Thus, if φ(x)=f^2 (x)⇒φ(−x)=−φ(x)=−f^2 (x) But,φ(−x)=f^2 (−x) and, ∀f(x)∈R, f^2 (−x)≥0. ⇒φ(−x)≥0⇒φ(−x)≮0 as implied by φ(−x)=−f^2 (x). ∴f^2 is not odd for any x∈R.⇒∄f∈R∣η(f)=0 if f(x)≠0. If we only consider the statement η(f)=0, we can have f(x)=0 being a function. If μ(f)=0⇒η(f)=0, 0+0=(M+m)∫_0 ^1 fdx−Mm ⇒∫_0 ^1 f(x)dx=((Mm)/(m+M))=((max(f)min(f))/(max(f)+min(f))) Is there a function f∈[m,M] such that ∫_0 ^1 f(x)dx=((mM)/(m+M))? By the fundamental theorem of calculus F(1)−F(0)=((max(F^′ (x))min(F^′ (x)))/(max(F^′ (x))+min(F^′ (x)))) If η(f)=0⇒μ(f)=(m+M)∫_0 ^1 fdx−Mm ∴ ∫_0 ^1 f^2 dx=(m+M)∫_0 ^1 fdx−mM ∫_0 ^1 (f^2 (x)−(m+M)f(x))dx=−mM ∫_0 ^1 f(x)(f(x)−m+M)dx=−mM The simpler equation to the above one is ∫_0 ^1 (M−f(x))(f(x)−m)dx=0. ∴ Let φ(x)=(M−f(x))(f(x)−m) where φ(x) is odd for x∈[0,1].
Anattemptmax(f)=M,min(f)=m∴f↑=M−f,f↓=f−mμ(f)=∫10f↑f↓dx=∫01(M−f)(f−m)dxμ(f)=∫01(M+m)fdx−∫01f2dx−Mm∫01dxμ(f)=(M+m)∫01fdx−η(f)−Mmμ(f)+η(f)=(M+m)∫01fdx−Mmf(x)=exη(ex)=∫01e2xdx=12(e2−1)=12(e+1)(e−1)∀x∈[0,1],ex∈[1,e]⇒m=1,M=e∴μ(ex)=(1+e)∫01exdx−12(e+1)(e−1)−eμ(ex)=(1+e)(e−1)−0.5(e+1)(e−1)−eμ(ex)=0.5(e+1)(e−1)−e≠0.5(e−1)(e+1)μ(ex)≠η(ex)f(x)=xη(x)=∫01x2dx=13∀x∈[0,1],f∈[0,1]⇒m=0,M=1.∴μ(x)=(0+1)∫01xdx−13−0μ(x)=12−13=16≠13∴μ(x)≠η(x)Forf(x)≠0∀x∈[0,1],ifη(f)=0⇒f2isoddforx∈[0,1]Thus,ifϕ(x)=f2(x)⇒ϕ(−x)=−ϕ(x)=−f2(x)But,ϕ(−x)=f2(−x)and,∀f(x)∈R,f2(−x)⩾0.⇒ϕ(−x)⩾0⇒ϕ(−x)≮0asimpliedbyϕ(−x)=−f2(x).∴f2isnotoddforanyx∈R.⇒∄f∈R∣η(f)=0iff(x)≠0.Ifweonlyconsiderthestatementη(f)=0,wecanhavef(x)=0beingafunction.Ifμ(f)=0⇒η(f)=0,0+0=(M+m)∫01fdx−Mm⇒∫01f(x)dx=Mmm+M=max(f)min(f)max(f)+min(f)Isthereafunctionf∈[m,M]suchthat∫01f(x)dx=mMm+M?BythefundamentaltheoremofcalculusF(1)−F(0)=max(F′(x))min(F′(x))max(F′(x))+min(F′(x))Ifη(f)=0⇒μ(f)=(m+M)∫01fdx−Mm∴∫01f2dx=(m+M)∫01fdx−mM∫01(f2(x)−(m+M)f(x))dx=−mM∫01f(x)(f(x)−m+M)dx=−mMThesimplerequationtotheaboveoneis∫01(M−f(x))(f(x)−m)dx=0.∴Letϕ(x)=(M−f(x))(f(x)−m)whereϕ(x)isoddforx∈[0,1].

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