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What are all continuous functions $f:[a,b] \to \mathbb{R}$ such that $\int_a^b f = sup_{[a,b]} f$

it's clear that it is not all constant functions as given $[0,1]$ of the function $f(x) = 1$ then we have equality but as soon as given an interval of magnitude greater than $1$ this will fail to be equality. For instance $[0,2]$ gives $2 \not = 1$. So if it doesn't hold for all constant functions then it cannot hold for all linear functions.

The only continuous function I can find that it works with is $f(x) = 0, \forall x \in \mathbb{R}$.

Am I missing something here? $a , b$ are independent of the function. I've been able to construct intervals which give equality for a given function but not universally of any $[a,b]$ other than the zero function.

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  • $\begingroup$ For $[0,2]$ it holds for any function of the form $ax$, for positive $a$. $\endgroup$ – Arthur Jan 29 '17 at 0:32
  • $\begingroup$ @Arthur but for [0,3] it would no longer be true. I don't get to pick what $a$ and $b$ are. $\endgroup$ – oliverjones Jan 29 '17 at 0:38
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The only continuous function $f$ such that $\int_a^b f(x) \, \mathrm{d}x = \sup_{[a,b]} f$ for all intervals $[a,b]$ is $f \equiv 0$. You can see this by letting $b = a + \varepsilon$ for arbitrary $a$ and $\varepsilon$ tending to zero: it follows that $$f(a) = \lim_{\varepsilon \rightarrow 0} \sup_{[a,a+\varepsilon]} f = \lim_{\varepsilon \rightarrow 0} \int_a^{a+\varepsilon} f(x) \, \mathrm{d}x = 0.$$

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