Basically, I have two statements, that should actually say the same thing, but one has a much stronger implication it.

One says, if $f: D \rightarrow \mathbb{R}$ is continuous and D is compact then $$ \exists x_{min},x_{max} \in D : \sup\limits_{x\in D} f(x) = f(x_{max}) \wedge \inf\limits_{x\in D} f(x) = f(x_{min}) $$

Which means that you can immediately deduce,

for $a<b$ $$f:[a,b] \rightarrow \mathbb{R}$$ the minimum is $f(a)$ and maximum is $f(b)$

Yet I'm not sure if that is true. Most of the time it is simply stated that $x_{min},x_{max} \in D$ exist, but not that you can immediately say, that $$\sup\limits_{x\in D} f(x) = f(x_{max})$$

Is the first statement true?

  • $\begingroup$ What you can "immediately deduce" is completely wrong, so nop: the first statement isn't true. $\endgroup$ – DonAntonio Feb 11 '17 at 12:59
  • $\begingroup$ Is the supremum of $f(x)$ not equivalent to the maximum of $f(x)$? $\endgroup$ – Jonathan Feb 11 '17 at 13:00
  • $\begingroup$ Yes sure, but why is the minimum in $a$? Have a look at $f(x)=-x$ and $D=[0,1]$. $\endgroup$ – Niklas Feb 11 '17 at 13:01
  • $\begingroup$ If the above statement is true, which I'm not sure of, I can find the minimal x value $x_{min}$ and $f(x_{min})$ should then be my infimum, which should be equivalent to the minimum. And in the case of $f:[a,b] \rightarrow \mathbb{R}$ , a is that minimal $x$ value, is it not? $\endgroup$ – Jonathan Feb 11 '17 at 13:04
  • $\begingroup$ @JonathanZ The minimal value of $\;x\in[a,b]\;$ is not necessarily the same as the minimal value of $\;f(x)\,,\,\,x\in[a,b]\;$ ! In fact, most probably it won't be. Take for example $\;f(x)=x(x-1)\;,\;\;x\in[0,1]\;$ , or the example Niklas gave you... $\endgroup$ – DonAntonio Feb 11 '17 at 13:09

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