It is given that $f$ is a continuous function and $A$ a bounded set.

Is the inequality $$\sup_{x \in A} \int_a^b f(x) dx \leq \int_a^b sup_{x \in A} f(x) dx$$ true?

If it is true,could we maybe prove it,with this relation: $$\sup_{x \in A} (\sup_{x \in A} f(x))=\sup_{x \in A} f(x)$$ ?

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    $\begingroup$ the expression on the left side is ill-formed. x can not be a variable of the sup and a variable of the integral at the same time. $\endgroup$ – Michael Apr 30 '14 at 19:08
  • $\begingroup$ Yes,right..but $\sup_{x \in A} f(x)$ is a real number,so it does not depend from $x$,or not? $\endgroup$ – evinda Apr 30 '14 at 19:10
  • $\begingroup$ @Michael It's not ill-formed, just weird. The integral operation re-binds $x$, i.e. introduces a new variable that happens to also be called $x$, and within the integral $x$ refers to that new variable. The outter variable $x$ - the one bound by the supremem - is therefore simply unused. $\endgroup$ – fgp Apr 30 '14 at 19:15
  • $\begingroup$ I think he meant $\sup_{a,b \in A}$ on the LHS $\endgroup$ – PA6OTA Apr 30 '14 at 19:20

The statement is false.


$$a=0, b=1, f(x)=x, A=(-3,-2)$$

Because: Who said that $A$ has to be the integration intervall?


As states, this is false. You need to impose some restriction on $a,b$ vs. $A$ for this to hold, otherwise you can do something like $a=0$, $b=2$, $A=[0,1]$ and let $f(x) = 0$ if $x \leq 1$ and $1$ otherwise. Then $$ \int_a^b f = 1 > \int_a^b \sup_{x\in[0,1]} f(x) = \int_a^b 0 = 0 \text{.} $$

Also, note that $x$ isn't a free variable in the expression $\int_a^b f(x)\,dx$, so writing $$ \sup_{x \in A} \int_a^b f(x)\,dx $$ is a bit weird. It comes down to writing $\sup_{x \in A} c$ for some constant $c$, which is of course just $c$.

However, you do have that $$ \int_A f(x) \,dx \leq \int_A |f(x)| \,dx \leq \int_A \sup_{x\in A}|f(x)| \,dx = |A|\cdot \sup_{x\in A}|f(x)| \text{,} $$ where $|A|$ denoted the measure of $A$, i.e. $|A| = \int_A 1 \,dx$.

You also have that if $f(x) < g(x)$ then $$ \int_A f(x) \,dx \leq \int_A g(x) \,dx \text{.} $$ For $f \geq 0$ this is clear. In the general case (this is where I was confused prevously), you can split $f$ into it's positive and negative parts, i.e. write $f(x) = f^+(x) - f^-(x)$ where $f^+,f^- \geq 0$, and the same for $g$. Then, $f^+ \leq g^+$ and $f^- \geq g^-$, and thus $$ \int_A g(x) - \int_A f(x) = \underbrace{\int_A g^+ - \int_A f^+}_{\geq 0} + \underbrace{\int_A f^- - \int_A g^-}_{\geq 0} > 0 $$

  • $\begingroup$ But f is continous... $\endgroup$ – rlartiga Apr 30 '14 at 19:14
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    $\begingroup$ Isn't it $$\sup_{x \in [0,1]} f(x) dx=0$$? $\endgroup$ – evinda Apr 30 '14 at 19:16
  • $\begingroup$ really... $\int_0^1 \sup_{x\in[0,1]} f(x) \,dx = 0$ !!! $\endgroup$ – Michael Apr 30 '14 at 19:20
  • $\begingroup$ @Michael Yeah, ups, sorry for that. $\endgroup$ – fgp Apr 30 '14 at 19:21

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