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Infimum of integral of open set

I have already done part a and b. Part a is easy, for part b, i let the anti-derivative of f to be k(t)+c and arrive at the answer that F(f)= 1/2+ 2*k(1/2) - k(1). But i don't know how to do the next part, can anyone give me a hint? the question c ask me to show that the infimum of F is 0 and it is never attained on A.

Re: Infimum of integral of open set

Hi kanezila,

In part b, to prove that $\displaystyle F$ is continuous at $\displaystyle f$ we need to show that for $\displaystyle \epsilon>0$ there is $\displaystyle \delta>0$ such that whenever $\displaystyle \|f-g\|_{\infty}<\delta$, we have $\displaystyle |F(f)-F(g)|<\epsilon.$ Perhaps looking at an antiderivative will go somewhere, but this is probably best done by using the definition.

To prove part c looking at the following sequence of functions $\displaystyle \{f_{n}\}$ in $\displaystyle A$, where $\displaystyle f_{n}(x)=0$ for $\displaystyle x\in [0,1/2]$, $\displaystyle f_{n}(x)=(n+2)(x-1/2)$ for $\displaystyle x\in [1/2, 1/2+1/(n+2)]$ and $\displaystyle f_{n}(x)=1$ for $\displaystyle x\in [1/2+1/(n+2), 1].$ Then we should be able to show that $\displaystyle F(f_{n})=\frac{1}{2}(\frac{1}{n+2}).$ Since this goes to 0 as n goes to infinity the infimum must be 0, because each of the $\displaystyle f_{n}$'s belong to $\displaystyle A.$

Does this clear things up? Good luck!

Re: Infimum of integral of open set

thank you very much for this, i have been trying to do this question for a whole day already.

Re: Infimum of integral of open set

could you tell me how to prove the first one???many thanks