Interchanging supremum and integral

Suppose that $\varphi$ is a smooth strictly increasing function with $\varphi(0)=0$ and $B$ is a compact subset of $R^{n}$. Let $x : R_{\geq 0} \times R^{n} \to R^{n}$ be such that $x(.,\xi)$ is differentiable for each $\xi$ and $x(t,.)$ is uniformly locally Lipschitz for each $t$ (i.e. there is some strictly positive constant $L$ such that $\| x(t,\xi_{1}) - x(t,\xi_{2})\| \leq L \| \xi_{1} - \xi_{2} \|$ for all $\xi_{1},\xi_{2} \in K \subset R^{n}$). I'd like to know that whether the following holds $$\sup_{\xi \in B} \int_{s=0}^{s=t}{\varphi(x(s,\xi))ds} = \int_{s=0}^{s=t}{\sup_{\xi \in B} \varphi(x(s,\xi))ds}.$$ In other word, are the supremum and integral signs interchangable?

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There is a type in the question. Please replace $R^{n}$ by $R$ – Navid Noroozi Dec 4 '12 at 6:59
There are typoes in the question. Please replace $R^{n}$ by $R$ and $x \colon R_{\geq 0} \times R \to R_{\geq 0} and$x(t,0) = 0$. – Navid Noroozi Dec 4 '12 at 7:26 2 Answers They're not interchangeable. Let$f$be a smooth, nonnegative function supported on$[0,1]$with$\int_0^1 f(x)dx = \frac{1}{2}$and$\sup f = 1$. Let$\phi(x) = x$and$x(s,\xi) = f(s - \xi)$for$\xi \in [0,1]$. The hypothesis on$\phi$is then satisfied, and$x(t,\xi) = f(t - \xi)$has a bounded partial derivative with respect to$\xi$, so the Lipschitz condition is satisfied.$x(t,\cdot) = f(t - \cdot)$is also differentiable since$f$is smooth. Now, the LHS of our desired equality is $$\sup_{\xi \in [0,1]} \int_0^2 f(s - \xi) ds = \frac{1}{2},$$ as translation will not change the value of the integral. On the other hand, suppose that the maximum of$f$occurs at$x_0$. Then for any$s \in [x_0, x_0+1]$, there is$\xi \in [0,1]$so that$f(s - \xi) = f(x_0) = 1$. Hence$\sup_{\xi \in [0,1]} f(s - \xi)$is 1 on the interval$[x_0, x_0+1]$. Hence the integral on the RHS must be at least 1. Edit: I made a little error. Uniform continuity does not imply Lipschitz! But the boundedness of$f'\$ will.

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What the questioner should make clear is what is the choice set on the left and right sides of the equality about. Let me write x for xi. On the left side it looks like you are assuming there is one choice of x for all s. On the right side it looks like you are assuming that for each s you can choose a different x(s). Obviously those are different.

It makes more sense to me to ask if you keep the choice set the same: the function (x(s): for all s) with whatever properties you want on x(s) . On the left side there will be indeterminacy of the choice because altering the value at individual points alone doesn't matter--because those points have zero measure- but the values of the integrals will be the same.

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