Let $f:\mathbb R\to\mathbb R$ have a continuous second derivative. Suppose for all $s,t \in \mathbb R$ with $s<t$ we have $$\frac1{t-s}\int_s^tf(x)\,dx=\frac{f(s)+f(t)}2.$$ Show there exist $\alpha$ and $\beta$ such that $f(x)=\alpha x+\beta$.

I think I use second fundamental theorem of calculus or I use the mean value theorem of integrals. Another thing I thought is the right hand side is the midpoint between two functions; so I think the function has to have the midpoint of the two endpoints involved.

  • $\begingroup$ Have you tried applying any of these ideas? $\endgroup$ – robjohn May 24 '18 at 14:49
  • $\begingroup$ Please provide a meaningful title next time your post a question. Requesting for help isn't providing good knowledge on your question! $\endgroup$ – mathcounterexamples.net May 24 '18 at 14:53
  • $\begingroup$ @robjohn So what I have is by the FTC, for any $p$ which $s \leq p \leq t$, then $\int_{s}^{x} f(p) dp = f(x)$. Then, $f(x) = \frac{d}{dx} (t - s) \frac{f(s) + f(t)}{2}$. Am I on the right track? $\endgroup$ – stevieman5933 May 24 '18 at 14:59
  • $\begingroup$ This is the integral version of the well-known series result: $\frac{1}{n-m+1}\sum_{k=m}^n a_k = \frac{a_m + a_n}{2}$ when $a_k$ is an arithmetic progression (a linear function). $\endgroup$ – Winther May 24 '18 at 15:09
  • $\begingroup$ @stevieman The FTC should read $\int_s^x f'(p)dp=f(x)-f(s)$ or $\partial_x\int_s^x f(p)dp=f(x)$. $\endgroup$ – J.G. May 24 '18 at 15:29

Since $\int_s^t fdx=(t-s)(f(t)+f(s))/2$, differentiating with respect to $t$ gives $f(t)=(f(t)+f(s))/2+(t-s)f'(t)/2$ so $f(t)-f(s)=(t-s)f'(t)$. Differentiating with respect to $s$ gives $-f'(s)=-f'(t)$, so $f'$ is constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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