# Inf sup of an expression not involving sequences

I have $$\inf_{f \in W} \sup_{x \in [-1,1]} |x + f(x)|$$ where $f \in W = \{f: [-1,1] \rightarrow R$ continuous $| \int_0^1 f(x) d \mu = \int_{-1}^0 f(x) d \mu = 0\}$.

I want to compute this thing but I'm not sure what to do with the $inf$ $sup$. I know that $\inf \sup x_n = \limsup x_n$ for sequences but here I don't have sequences. Does anyone have any idea about what I could do?

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Yes, that's what I meant. You are saying $\inf \sup |x + f| = \inf || x - f ||_\infty$ where the inf is taken over all $f \in W$? – Rudy the Reindeer Sep 30 '11 at 18:08
Did you mean $\| x -f(x) \|$ or $\| x+f(x) \|$? I am guessing that in this question, the two expressions give the same answer (since you can consider $-f$ instead of $f$), but still you should clarify. – Srivatsan Sep 30 '11 at 18:20
I mean $+$ as I wrote it in the question. – Rudy the Reindeer Sep 30 '11 at 18:27

Often we can ignore continuity at first and get functions that need to be approximated by continuous ones.

Consider $g(x)=x+f(x)$. We want to find $$\inf_{g \in W'} \sup_{x \in [-1,1]} |g(x)|\tag{1}$$ where $W'=\{g:[-1,1]\mapsto\mathbb{R}|\int_0^1g(x)\;\mathrm{d}x=\frac{1}{2}\text{ and }\int_{-1}^0g(x)\;\mathrm{d}x=-\frac{1}{2}\}$.

In $(1)$ we want to find a $g$ that minimizes the supremum of its absolute value, yet has integral $\frac{1}{2}$ on $[0,1]$ and integral $-\frac{1}{2}$ on $[-1,0]$. On $[0,1]$, this would be $g(x)=\frac{1}{2}$ and on $[-1,0]$, it would be $g(x)=-\frac{1}{2}$. Thus, we get that $$f(x)=\left\{\begin{array}{}+\frac{1}{2}-x&x\in[0,+1]\\-\frac{1}{2}-x&x\in[-1,0]\end{array}\right.\tag{2}$$ Now, $(2)$ doesn't describe a continuous function, but it can easily be approximated by one that has behavior as close to $f$ as we wish. Thus, the answer to the original question is $\frac{1}{2}$.

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Thank you Davide and robjohn, I merged your answers into the following:

To compute the following: $$|| f||_{C/W} = || x||_{C/W} = \inf_{f \in W} \sup_{x \in [-1,1]} |x + f(x)|$$

Let $g(x) = x + f(x)$. Then $$\int_0^1 g(x) d \mu = \frac{1}{2}$$ and $$\int_{-1}^0 g(x) d \mu = - \frac{1}{2}$$

$$\implies || f||_{C/W} = \inf_{g \in W^\prime} \sup_{x \in [-1,1]} |g(x)| = \inf_{g \in W^\prime} ||g ||_\infty$$ where $W^\prime = \{ g: [-1,1] \rightarrow R | \int_0^1 g(x) d \mu = \frac{1}{2}, \int_{-1}^0 g(x) d \mu = - \frac{1}{2}\}$

(i) $$|| g||_\infty = || g||_\infty \mu([0,1]) \geq \int_0^1 g d \mu = \frac{1}{2}$$

$$\implies ||f||_{C/W} \geq \frac{1}{2}$$

(ii) Now we want a function in $W$ whose supremum (=maximum) is $\frac{1}{2}$. That's easy:

$$f(x) := -x - \frac{1}{2}$$ for $x \in [-1,0]$ $$f(x) := x - \frac{1}{2}$$ for $x \in [0,1]$

$$\implies || f ||_{C/W} = \frac{1}{2}$$

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