Welcome to our community

Be a part of something great, join today!

Is f bounded??

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hey!!! :)
I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.
Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??
 

chisigma

Well-known member
Feb 13, 2012
1,704
Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??
If $f:[a,b]$ is Riemann integrable, then necessarly f(*) is bounded in [a,b]...


Kind regards


$\chi$ $\sigma$
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??
Hello.

There are two ways to define what it means to be "integrable" on $[a,b]$.
(i) The original Riemann definition: there exists a real number $I$ such that for any $\varepsilon > 0$, there is a $\delta > 0$ so that for any partition with norm $<\delta$ we have that all Riemann sums with respect to partition are within $\varepsilon$ of $I$.
(ii) The modified Darboux definition: the lower integral and the upper integral are equal.

In (i) there is no mention of boundness. In (ii) it is assumed that the function is bounded because we work with its sup and inf. But it is not hard to show that in (i) boundness is a necessary condition of an integrable function.

Which definition is it that you use?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Hey!!! :)
I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.
Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??
Hello.

There are two ways to define what it means to be "integrable" on $[a,b]$.
(i) The original Riemann definition: there exists a real number $I$ such that for any $\varepsilon > 0$, there is a $\delta > 0$ so that for any partition with norm $<\delta$ we have that all Riemann sums with respect to partition are within $\varepsilon$ of $I$.
(ii) The modified Darboux definition: the lower integral and the upper integral are equal.

In (i) there is no mention of boundness. In (ii) it is assumed that the function is bounded because we work with its sup and inf. But it is not hard to show that in (i) boundness is a necessary condition of an integrable function.

Which definition is it that you use?
Alternatively, you could use the Riemann-Lebesgue Theorem, if you've gotten that far and are allowed to use it.
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Alternatively, you could use the Riemann-Lebesgue Theorem, if you've gotten that far and are allowed to use it.
I doubt you mean the one from Fourier analysis. So you probably mean the one that says a function is integrable if and only if its set of discontinuity is of measure zero. But this theorem says nothing about boundedness.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
I doubt you mean the one from Fourier analysis.
Right - I've usually heard that one called the "Riemann-Lebesgue Lemma".

So you probably mean the one that says a function is integrable if and only if its set of discontinuity is of measure zero.
Right.

But this theorem says nothing about boundedness.
I agree, but it would fairly straight-forwardly prove that if $f$ is integrable, then $f^2$ is integrable. Boundedness requires some other machinery.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Hey!!! :)
I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.
Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??
If $f$ is the function defined on the unit interval [0,1] by $f(x) = x^{-1/2}$ for $x>0$, and $f(0) = 0$, then \(\displaystyle \int_0^1f(x)\,dx\) exists as an improper Riemann integral, but $f^2$ is not integrable.

The definition of the Riemann integral requires that the function should be bounded. In order to deal with unbounded functions you need to use the improper Riemann integral. In that situation it is not true that the square of an integrable function is integrable, as the above example shows. Presumably your question refers to the ordinary Riemann integral for bounded functions.