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???](Q2032.png)
$${f}:\left[\mathrm{0},+\mathrm{1}\right]\rightarrow\mathbb{R} \\ $$$$\eta\left({f}\right):=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}^{\mathrm{2}} {dx} \\ $$$$\mathrm{and}\:\mathrm{if}\:\forall{x}\in\left[\mathrm{0},\mathrm{1}\right],{f}\in\left[{m},\mathrm{M}\right]\:\mathrm{for}\:\mathrm{some}\:\left({m},\mathrm{M}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$${f}\uparrow:=\mathrm{max}\left({f}\right)−{f} \\ $$$${f}\downarrow:={f}−\mathrm{min}\left({f}\right) \\ $$$$\mu\left({f}\right):=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\downarrow{f}\uparrow{dx} \\ $$$$\mathrm{then} \\ $$$${f}={e}^{{x}} ,\:\eta\left({f}\right)\overset{?} {=}\mu\left({f}\right) \\ $$$${f}={x},\eta\left({f}\right)\overset{?} {=}\mu\left({f}\right) \\ $$$$\exists{f},\eta\left({f}\right)=\mathrm{0}??? \\ $$$$\exists{f},\mu\left({f}\right)=\mathrm{0}??? \\ $$$$\mu\left({f}\right)=\mathrm{0},\:\mathrm{does}\:\eta\left({f}\right)=\mathrm{0}??? \\ $$$$\eta\left({f}\right)=\mathrm{0},\:\mathrm{does}\:\mu\left({f}\right)=\mathrm{0}??? \\ $$
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].](Q2033.png)
$${An}\:{attempt} \\ $$$${max}\left({f}\right)={M},{min}\left({f}\right)={m} \\ $$$$\therefore\:{f}\uparrow={M}−{f}\:,\:{f}\downarrow={f}−{m} \\ $$$$\mu\left({f}\right)=\overset{\mathrm{1}} {\int}_{\mathrm{0}} {f}\uparrow{f}\downarrow{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({M}−{f}\right)\left({f}−{m}\right){dx} \\ $$$$\mu\left({f}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left({M}+{m}\right){fdx}−\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} {dx}−{Mm}\int_{\mathrm{0}} ^{\mathrm{1}} {dx} \\ $$$$\mu\left({f}\right)=\left({M}+{m}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {fdx}−\eta\left({f}\right)−{Mm} \\ $$$$\mu\left({f}\right)+\eta\left({f}\right)=\left({M}+{m}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {fdx}−{Mm} \\ $$$${f}\left({x}\right)={e}^{{x}} \\ $$$$\eta\left({e}^{{x}} \right)=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\mathrm{2}{x}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}} −\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({e}+\mathrm{1}\right)\left({e}−\mathrm{1}\right) \\ $$$$\forall{x}\in\left[\mathrm{0},\mathrm{1}\right],{e}^{{x}} \in\left[\mathrm{1},{e}\right]\Rightarrow{m}=\mathrm{1},{M}={e} \\ $$$$\therefore\mu\left({e}^{{x}} \right)=\left(\mathrm{1}+{e}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}} {dx}−\frac{\mathrm{1}}{\mathrm{2}}\left({e}+\mathrm{1}\right)\left({e}−\mathrm{1}\right)−{e} \\ $$$$\mu\left({e}^{{x}} \right)=\left(\mathrm{1}+{e}\right)\left({e}−\mathrm{1}\right)−\mathrm{0}.\mathrm{5}\left({e}+\mathrm{1}\right)\left({e}−\mathrm{1}\right)−{e} \\ $$$$\mu\left({e}^{{x}} \right)=\mathrm{0}.\mathrm{5}\left({e}+\mathrm{1}\right)\left({e}−\mathrm{1}\right)−{e}\neq\mathrm{0}.\mathrm{5}\left({e}−\mathrm{1}\right)\left({e}+\mathrm{1}\right) \\ $$$$\mu\left({e}^{{x}} \right)\neq\eta\left({e}^{{x}} \right) \\ $$$$ \\ $$$${f}\left({x}\right)={x} \\ $$$$\eta\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\forall{x}\in\left[\mathrm{0},\mathrm{1}\right],\:{f}\in\left[\mathrm{0},\mathrm{1}\right]\Rightarrow{m}=\mathrm{0},{M}=\mathrm{1}. \\ $$$$\therefore\mu\left({x}\right)=\left(\mathrm{0}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {xdx}−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{0} \\ $$$$\mu\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}}\neq\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore\mu\left({x}\right)\neq\eta\left({x}\right) \\ $$$$ \\ $$$${For}\:{f}\left({x}\right)\neq\mathrm{0}\:\forall{x}\in\left[\mathrm{0},\mathrm{1}\right],\:{if}\:\eta\left({f}\right)=\mathrm{0}\Rightarrow{f}^{\mathrm{2}} \:{is}\:{odd}\:{for}\:{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${Thus},\:{if}\:\phi\left({x}\right)={f}^{\mathrm{2}} \left({x}\right)\Rightarrow\phi\left(−{x}\right)=−\phi\left({x}\right)=−{f}^{\mathrm{2}} \left({x}\right) \\ $$$${But},\phi\left(−{x}\right)={f}^{\mathrm{2}} \left(−{x}\right)\:{and},\:\forall{f}\left({x}\right)\in\mathbb{R},\:{f}^{\mathrm{2}} \left(−{x}\right)\geqslant\mathrm{0}. \\ $$$$\Rightarrow\phi\left(−{x}\right)\geqslant\mathrm{0}\Rightarrow\phi\left(−{x}\right)\nless\mathrm{0}\:{as}\:{implied}\:{by} \\ $$$$\phi\left(−{x}\right)=−{f}^{\mathrm{2}} \left({x}\right).\:\:\therefore{f}^{\mathrm{2}} \:{is}\:{not}\:{odd}\:{for}\:{any} \\ $$$${x}\in\mathbb{R}.\Rightarrow\nexists{f}\in\mathbb{R}\mid\eta\left({f}\right)=\mathrm{0}\:{if}\:{f}\left({x}\right)\neq\mathrm{0}. \\ $$$${If}\:{we}\:{only}\:{consider}\:{the}\:{statement} \\ $$$$\eta\left({f}\right)=\mathrm{0},\:{we}\:{can}\:{have}\:{f}\left({x}\right)=\mathrm{0}\:{being}\:{a} \\ $$$${function}. \\ $$$$ \\ $$$${If}\:\mu\left({f}\right)=\mathrm{0}\Rightarrow\eta\left({f}\right)=\mathrm{0},\:\mathrm{0}+\mathrm{0}=\left({M}+{m}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {fdx}−{Mm} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\frac{{Mm}}{{m}+{M}}=\frac{{max}\left({f}\right){min}\left({f}\right)}{{max}\left({f}\right)+{min}\left({f}\right)} \\ $$$${Is}\:{there}\:{a}\:{function}\:{f}\in\left[{m},{M}\right]\:{such}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\frac{{mM}}{{m}+{M}}? \\ $$$${By}\:{the}\:{fundamental}\:{theorem}\:{of}\:{calculus} \\ $$$$\:\:{F}\left(\mathrm{1}\right)−{F}\left(\mathrm{0}\right)=\frac{{max}\left({F}^{'} \left({x}\right)\right){min}\left({F}^{'} \left({x}\right)\right)}{{max}\left({F}^{'} \left({x}\right)\right)+{min}\left({F}^{'} \left({x}\right)\right)} \\ $$$${If}\:\eta\left({f}\right)=\mathrm{0}\Rightarrow\mu\left({f}\right)=\left({m}+{M}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {fdx}−{Mm} \\ $$$$ \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} {dx}=\left({m}+{M}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {fdx}−{mM} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}^{\mathrm{2}} \left({x}\right)−\left({m}+{M}\right){f}\left({x}\right)\right){dx}=−{mM} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\left({f}\left({x}\right)−{m}+{M}\right){dx}=−{mM} \\ $$$${The}\:{simpler}\:{equation}\:{to}\:{the}\:{above}\:{one} \\ $$$${is}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({M}−{f}\left({x}\right)\right)\left({f}\left({x}\right)−{m}\right){dx}=\mathrm{0}. \\ $$$$\therefore\:{Let}\:\phi\left({x}\right)=\left({M}−{f}\left({x}\right)\right)\left({f}\left({x}\right)−{m}\right) \\ $$$${where}\:\phi\left({x}\right)\:{is}\:{odd}\:{for}\:{x}\in\left[\mathrm{0},\mathrm{1}\right]. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$