12

Prove this:

Let $f :[a,b] \to \mathbb{R}$ be a bounded and integrable function. Show that $f^2$ is integrable too.

I'm in trouble with this. Can anyone show how to do it?

Hans Lundmark
  • 55,038
emma
  • 297
  • 2
    What do you mean f is limited. Does it mean that $\exists M\in\mathbb{R}$ such that $|f(x)|\le M$ for all $x\in\mathbb{R}$? Is this with respect to the Riemann integral or the Lebesgue integral? – Laars Helenius Oct 13 '14 at 22:25
  • 1
    I changed “limited” to ”bounded”, which is the most commonly used term in English. – egreg Oct 13 '14 at 23:08

3 Answers3

21

Considering Riemann integrals, you can make the following argument.

Notice that if $f$ is bounded with $|f(x)| \leq B$, then

$$|f^2(x) - f^2(y)|= |f(x)-f(y)||f(x)+f(y)|\leq 2B|f(x)-f(y)|$$

and

$$M_j(f^2)-m_j(f^2)=\sup \{f^2(x):x_{j-1}\leq x\leq x_j\}-\inf \{f^2(x):x_{j-1}\leq x\leq x_j\}\leq 2B[M_j(f)-m_j(f)]$$

The difference of upper and lower sums for $f^2$ with respect to a partition $P$ is

$$U(P,f^2) - L(P,f^2) = \sum_{j=1}^{n}[M_j(f^2)-m_j(f^2)](x_j-x_{j-1}) \leq 2B[U(P,f) - L(P,f)].$$

Since $f$ is integrable, for any $\epsilon > 0$ there exists a partition P such that

$$U(P,f) - L(P,f) < \frac{\epsilon}{2B},$$

and

$$U(P,f^2) - L(P,f^2) < \epsilon.$$

Therefore, $f^2$ satisfies the Riemann criterion for integrability.

RRL
  • 92,835
  • 7
  • 70
  • 142
  • 1
    How do you claim $ M_j(f^2)-m_j(f^2)=\sup {f^2(x):x_{j-1}\leq x\leq x_j}-\inf {f^2(x):x_{j-1}\leq x\leq x_j}\leq 2B[M_j(f)-m_j(f)] $ ? Because sup of any function does not necessarily have to belong to the range of the function. – L lawliet Nov 13 '23 at 17:52
  • As $f$ and $f^2$ are bounded the suprema and infima are finite. Also true -- $f$ is integrable but it is not necessarily continuous everywhere and the supremum and/or infimum are not necessarily attained on subintervals. But what makes you think I am assuming that they are attained? – RRL Nov 13 '23 at 20:21
  • @L lawliet: Note that $M_j(f^2) := \sup{f^2(x): x_{j-1}\leqslant x \leqslant x_j}$ is just the definition for the $M_j$ notation. I'm not saying that $M_j(f^2)$ is attained by f^2 at any point in the interval. – RRL Nov 13 '23 at 23:19
  • if they are not attained then how can you plug them in the equation $ |f^2(x) - f^2(y)|= |f(x)-f(y)||f(x)+f(y)|\leq 2B|f(x)-f(y)| $ , because that is what I thought was done in this proof ? – L lawliet Nov 18 '23 at 14:56
  • @Llawliet: Read about Darboux sums here. – RRL Nov 20 '23 at 17:56
  • 2
    Starting with $|f^2(x) - f^2(y)|= |f(x)-f(y)||f(x)+f(y)|\leq 2B|f(x)-f(y)|$ with $x,y \in [x_{j-1},x_j$, the factor $|f(x) - f(y)|$ on the RHS is bounded by $\sup_{x,y\in [x_{j-1},x_j]} |f(x) - f(y)|$. That can be shown to equal $\sup_{z\in[x_{j-1},x_j]} f(z) - \inf_{z\in[x_{j-1},x_j] }f(z) := M_j(f) - m_j(f)$. The $M_j(f)$ and $m_j(f)$ is just standard shorthand notation for the supremum and infimum of $f$ on the interval, respectively. I make no assumption that $f$ attains these values at any points in the interval. (That will be true though if $f$ is continuous. ) – RRL Nov 20 '23 at 18:10
  • 1
    Now you have for all $x,y \in [x_{j-1},x_j]$ that $|f^2(x) - f^2(y)|\leq 2B [M_j(f) - m_j(f)]$. Take the supremum of the LHS to get $M_j(f^2) - m_j(f^2) \leq 2B[M_j(f) - m_j(f)]$ – RRL Nov 20 '23 at 18:14
7

If you are familiar with it, this follows immediately from Lebesgue Integrability Criteria.

As $f$ is bounded, so is $f^2$. Moreover, the set of discontinuities of $f^2$ is a subset of the set of discontinuities of $f$.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
N. S.
  • 134,609
  • how we can conclude that the set of discontinuities of $f^2$ is a subset the set of discontinities of $f$ ? – Micheal Brain Hurts Oct 29 '17 at 16:49
  • @Photoneaterman This is equivalent to proving that if $f$ is continuous at $a$ then $f^2$ is continuous at $a$. – N. S. Oct 29 '17 at 18:12
  • but you've said that " subset " so I wonder how to find it. For example if we took a function (1 for rationals, -1 for irrationals) it is nowhere continuous but $f^2$ is everywhere continuous, as you've written the answer of yours, " the set of discontinuities of $f^2$ is a subset of the set of discontinuities of $f$" is being true. – Micheal Brain Hurts Oct 29 '17 at 18:22
  • @Photoneaterman You don't "find" it. The statement "if $f$ is continuous at $a$ then $f^2$ is continuous at $a$" is equivalent to "if $f^2$ is discontinuous at $a$ then $f$ is discontinuous at $a$". This means exactly that the set of discontinuities is a subset. – N. S. Oct 29 '17 at 18:45
0

Composition theorem states that if

$f\in \mathcal{R}([a,b])$ with $f([a,b])\in[c,d]$ and $\phi:[c,d]\to\mathbb{R}$ then $\phi(f)\in\mathcal{R}([a,b])$

where $f\in \mathcal{R}([a,b])$ denotes that a function is reimann integrable over the closed interval $[a,b]$

let, $\phi=x^2$ and we are done

Nucleo
  • 660