give an example

**flower3** · February 16th 2010, 09:05 AM

Give an example of a function $f: [0,1] \to R$ such that $f \in R[0,1]$ [i.e f is Riemann integrable over [0,1]] $, f(x)>0, \ \forall x \in [0,1] , \ but \ \frac{1}{f}$ is not in $R [0,1]$

**tonio** · February 16th 2010, 09:12 AM

Originally Posted by flower3

Give an example of a function $f: [0,1] \to R$ such that $f \in R[0,1]$ [i.e f is Riemann integrable over [0,1]] $, f(x)>0, \ \forall x \in [0,1] , \ but \ \frac{1}{f}$ is not in $R [0,1]$

$f(x)=\left\{\begin{array}{ll}1&\;\;if\,\,x=0\\x&\; \;otherwise\end{array}\right.$

Tonio

**Drexel28** · February 16th 2010, 12:45 PM

Originally Posted by flower3

Give an example of a function $f: [0,1] \to R$ such that $f \in R[0,1]$ [i.e f is Riemann integrable over [0,1]] $, f(x)>0, \ \forall x \in [0,1] , \ but \ \frac{1}{f}$ is not in $R [0,1]$

tonio's answer is of course great, but more generally all you need to do is find a function which has one discontinuity on $[0,1]$ (guess where) such that it's multiplicative inverse is unbounded!

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