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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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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