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So <math>f(x)= f(0) \ \forall x\in[0,a]</math> and <math>f(x)= f(0)-1 \ \forall x\in(a,1]</math>. Thus | So <math>f(x)= f(0) \ \forall x\in[0,a]</math> and <math>f(x)= f(0)-1 \ \forall x\in(a,1]</math>. Thus | ||
| − | <math>\frac{1}{3}=\int_0^1f(x)dx=af(0) + (1-a)(f(0)-1)</math> | + | <math>\frac{1}{3}=\int_0^1f(x)dx=af(0) + (1-a)(f(0)-1)</math> and so, <math>\frac{4}{3}=a+f(0)</math>. |
| + | |||
| + | Now, given, <math>a \in [0,1]</math>, let <math>b=\frac{4}{3}-a</math>, we have | ||
| + | |||
| + | <math>f_a(x)=b \ \forall x\in [0,a]</math> and | ||
| + | <math>f_a(x)=b-1 \ \forall x\in (a,1]</math> | ||
Latest revision as of 09:39, 22 July 2008
From the identity $ f(0)-(V_{0}^{x})^{1/2} = f(x) $ $ \forall x\in[0,1] $ we notice that $ V $ is a positive and increasing function, therefore, $ f $ is decreasing. Hence $ f(x)-f(0)=-V_{0}^{x}) $.
We then have $ V_{0}^{x}=(V_{0}^{x})^{2} $
It means that there is a point $ a $ in $ [0,1] $ such that $ V $ jumps from $ 0 $ to $ 1 $ right after the point. (It has to occur like that in order to fulfill the identity.)
So $ f(x)= f(0) \ \forall x\in[0,a] $ and $ f(x)= f(0)-1 \ \forall x\in(a,1] $. Thus
$ \frac{1}{3}=\int_0^1f(x)dx=af(0) + (1-a)(f(0)-1) $ and so, $ \frac{4}{3}=a+f(0) $.
Now, given, $ a \in [0,1] $, let $ b=\frac{4}{3}-a $, we have
$ f_a(x)=b \ \forall x\in [0,a] $ and $ f_a(x)=b-1 \ \forall x\in (a,1] $
