1
$\begingroup$

On Wikipedia, when Proving a generalization of the mean value theorem, they state "Suppose f : [a, b] → R is continuous and g is a nonnegative integrable function on [a, b]. By the extreme value theorem, there exists m and M such that for each x in [a, b], $ {\displaystyle m\leqslant f(x)\leqslant M} $and $ {\displaystyle f[a,b]=[m,M]}$. They then proceed to make several evaluations. The part I am confused about is the final statement: $${\displaystyle f[a,b]=[m,M]}$$ What is this intended to mean( it was pointed out what I previously said here was incorrect, and I couldn't find the strike option, so I removed and replaced it). I couldn't find anything obvious by searching google, so I Decided to ask here. If this is the wrong place to ask this, please move it to the right place. Thanks!

( The proof is located at https://en.m.wikipedia.org/wiki/Mean_value_theorem under MVT for definite integrals, the first proof )

$\endgroup$
  • $\begingroup$ It certainly doesn't mean anything of the form "$\forall m,M\,\dots$" because it explicitly says "there exists $m$ and $M$". $\endgroup$ – Andreas Blass Feb 19 '18 at 21:52
  • $\begingroup$ Yeah after you pointed that out I re-read it and realized it was nonsensical. Couldn't find a strike option so I removed it. Thanks $\endgroup$ – Shinaolord Feb 19 '18 at 21:57
2
$\begingroup$

It means that the image of the interval $[a, b]$ under the function $f$ is the interval $[m, M]$. I.e., for every $x \in [a, b]$, $f(x) \in [m, M]$ and for every $y \in [m, M]$ there is an $x \in [a, b]$ such that $f(x) = y$. If $f$ is a function and $X$ is a set it is standard to use $f(X)$ to mean image of $X$ under $f$, i.e., the set $\{y \mid \exists x \in X(f(x) = y)\}$. The Wikipedia article is using something like this notation but writing $f[a, b]$ rather than $f([a, b])$. (Many writers, myself included, prefer to write $f[X]$ for the image, so we'd write $f[[a, b]]$ in this case.)

$\endgroup$
  • $\begingroup$ So by EVT, $\exists m,M$ such that $m\leq f(x) \leq M\ \forall x \in [a,b]$, and by the intermediate value theorem, since f(x) can take on all values in [m,M] since f is continuous, Its image is the interval. $\endgroup$ – Shinaolord Feb 19 '18 at 21:49
  • $\begingroup$ The previous statement should end with a question mark, my mistake. $\endgroup$ – Shinaolord Feb 19 '18 at 21:59
  • $\begingroup$ Your comment works without the question mark too! (I.e., the answer to the question is "yes".) $\endgroup$ – Rob Arthan Feb 19 '18 at 22:35
  • $\begingroup$ Thanks, I wasn't 100% on its validity so I figured it'd be smart to state a question mark was intended :). The notation I was taught would be $Im\lbrace f(x), \ x\in [a,b]\rbrace= [m,M]$ $\endgroup$ – Shinaolord Feb 19 '18 at 22:38
0
$\begingroup$

It means that for all $y$ such that $y \in [m,M]$ there is an $x \in [a,b]$ such that $f(x)=y$. You are not guaranteed that $f$ is constant on any interval. For example, we could have $f(x)=2x, [a,b]=3,4$. Then $m=6, M=8$ and for any $y \in [6,8]$ there is an $x \in [3,4]$ such that $2x=y$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.