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Consider $$ L(x):= \lim_{\epsilon \downarrow 0} \frac{\int_{x}^{x + \epsilon}f(t)dt}{\epsilon} $$

Is $L(x) = f(x)$ for any function $f$ that admits a Riemann integral? If not, are there any necessary and/or sufficient conditions on $f$ for this to happen?

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    If I am not mistaken, it is one of the Fundamental Theorem of Calculus. – Jo' Oct 14 '18 at 07:12
  • Could you please explain exactly how? I'm looking at both the FTCs from wikipedia but cannot exactly associate. https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements – sixtyTonneAngel Oct 14 '18 at 07:22
  • iirc, a function $f$ is Reiman integrable iff it is continuous almost everywhere. That would seem to imply $L(x)=f(x)$ almost everywhere... But analysis is not my strong suit.... – Xiaomi Oct 14 '18 at 07:31
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    RHS here is the derivative of $F(x) = \int_a^x f(t) ,dt$, so from Fundamental theorem of Calculus, any continuous function will do – Jakobian Oct 14 '18 at 07:48
  • But continuity is not a necessary condition on $f$ to admit a Riemann integral. Will the equality hold for all functions $f$ that admit a Riemann integral? – sixtyTonneAngel Oct 14 '18 at 08:06
  • $F$ doesn't have to be differentiable everywhere (i. e. $L$ doesn't have to exists for all $x$). But, there's a theorem, that tells us that $F$ is differentiable almost everywhere, and the equality $f(x)=F'(x)$ holds almost everywhere – Jakobian Oct 14 '18 at 09:29
  • If $f$ is continuous at $x$ then $L(x) =f(x) $ but continuity is only a sufficient and not necessary condition for this to hold. For example if $f(x) =\cos (1/x), f(0)=0$ then $L(0)=f(0)=0 $ but $f$ is not continuous at $0$. – Paramanand Singh Oct 14 '18 at 09:56
  • @ParamanandSingh : could you please add a computation showing how you arrived at $L(0)$ in your case? – sixtyTonneAngel Oct 14 '18 at 10:06
  • See https://math.stackexchange.com/a/1784625/72031 A more complicated example is available at https://math.stackexchange.com/q/2063078/72031 – Paramanand Singh Oct 14 '18 at 10:11

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