f integrable => f^2 integrable?

**Anonymous1** · November 6th 2009, 07:35 PM

Here is one that I don't really know how to approach:

Prove: if $f(x)$ is integrable on $[a,b],$ then $(f(x)^2)$ is integrable on $[a,b].$

**Enrique2** · November 7th 2009, 03:39 AM

This is not true. $f(x)=\frac{1}{\sqrt{x}}$ on [0,1] provides a counterexample. Perhaps you have been asked to prove the converse.

**Plato** · November 7th 2009, 05:05 AM

Originally Posted by Enrique2

This is not true. $f(x)=\frac{1}{\sqrt{x}}$ on [0,1] provides a counterexample.

In fact that is not a counterexample. $f(x)=\frac{1}{\sqrt{x}}$ is not Riemann integrable on $[0,1]$ .
Note that the function is not bounded on $[0,1]$ .

This happens to be a standard theorem/problem in the study Riemann integrals.
Here is the standard hint: $f(x)^2 - f(y)^2 = \left( {f(x) + f(y)} \right)\left( {f(x) - f(y)} \right)$ .
Use that to show that $U\left[ {f^2 ;P} \right] - L\left[ {f^2 ;P} \right] \leqslant 2 B\left( {U\left[ {f;P} \right] - L\left[ {f;P} \right]} \right)$ where $B$ is the bound of $f$ on $[a,b]$

**Enrique2** · November 7th 2009, 05:31 AM

Originally Posted by Plato

In fact that is not a counterexample. $f(x)=\frac{1}{\sqrt{x}}$ is not Riemann integrable on $[0,1]$ .
Note that the function is not bounded on $[0,1]$ .

This happens to be a standard theorem/problem in the study Riemann integrals.
Here is the standard hint: $f(x)^2 - f(y)^2 = \left( {f(x) + f(y)} \right)\left( {f(x) - f(y)} \right)$ .
Use that to show that $U\left[ {f^2 ;P} \right] - L\left[ {f^2 ;P} \right] \leqslant 2 B\left( {U\left[ {f;P} \right] - L\left[ {f;P} \right]} \right)$ where $B$ is the bound of $f$ on $[a,b]$

OK, sorry. Sure that the question is about Riemann integrability. I related the question to $L^p$ spaces.

**Anonymous1** · November 7th 2009, 01:12 PM

Since it is bounded, can I just assume that there exists a partition $P$ of $[a,b]$ such that

$U_{f}(P) - L_{f}(P) < \frac{\epsilon}{2B}?$

**Plato** · November 7th 2009, 01:40 PM

Originally Posted by Anonymous1

Since it is bounded, can I just assume that there exists a partition $P$ of $[a,b]$ such that

$U_{f}(P) - L_{f}(P) < \frac{\epsilon}{2B}?$

That is the idea because you know that $f$ is integrable.

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