Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.