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 )

  • $\begingroup$ 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

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.)

  • $\begingroup$ 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

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 $2x=y$


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