Let be $f:I\to\mathbb{R}$ a continuous function, $[x,x_0]\subseteq I$ and $M:=\sup\{f(y)\mid y\in[x,x_0]\}$. Show that the supremum attains the value of $f(x_0)$ when $x\to x_0$.

My original idea was to argue with the rules of limits:

By definition of the supremum we know that for each $\epsilon>0$ there exists a $y\in [x,x_0]$ such that $f(y)<M<f(y)+\epsilon$. Then $x\to x_0$ leads to $$\implies y\to x_0\implies\lim\limits_{y\to x_0}f(y)=f(x_0)\\\implies f(x_0)<M<f(x_0)+\epsilon \implies M=f(x_0),$$

as we can make $\epsilon>0$ arbitrarily small.

However, when I submitted this answer my tutor put a comment saying that this conclusion is not completely wrong (and maybe intuitively right) but not quite rigorous. Further he asked which specific rule would allow me to do that and finally deducted points.

Second attempt with $\epsilon$-$\delta$-criterion:

Again by definition of the supremum we know that for each $\epsilon>0$ there exists a $y\in [x,x_0]$ such that $f(y)<M<f(y)+\epsilon$. So if I now consider $x\to x_0$ then for an arbitrarily small $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$. As $y\in [x,x_0]$ it follows $f(x_0)-\epsilon<f(y)<M<f(y)<f(x_0)+\epsilon$. Hence $M=f(x_0)$.

I don't understand what the problem was with my first attempt? Is my second one right?


Beside my both approaches I am wondering if I am able to simply argue $\lim\limits_{x\to x_0}M=\sup\{f(y)\mid y\in[x,x_0]\}=\sup\{f(y)\mid y\in[x_0,x_0]\}=f(x_0)$ because if I shrink the interval $[x,x_0]$ then $x_0$ is the only value that's left. However in this approach continuity doesn't play a role which seems strange!?

  • 1
    $\begingroup$ Plural of supremum is suprema $\endgroup$ Nov 25 '20 at 15:17
  • $\begingroup$ I think one reason for your difficulties is that you treat $M$ as if it were constant, and it is not. It is really $M(x)$, and what you are trying to prove is that $\lim_{x\to x_0}M(x)=f(x_0)$. I don't know what theorems you have at your disposal, but as soon as you have the theorem that a continuous function on a closed and bounded interval is bounded and attains its bounds then this result is straightforward. $\endgroup$ Nov 25 '20 at 21:04

The problem in your first attempt is that, $y$ depends on $\varepsilon .$ Roughly speaking, you can regard this dependence as a one-to-one correspondence(N.B. this saying may not be logically correct because some $y$ may corresponds to many different $\varepsilon .$ But what I want to say is that, when you change $y$, $\varepsilon$ may be changed too.) So, when you send $y$ to $x_0$, $\varepsilon$ may change too. And it's quite hard to manage two changes simultaneously. And actually the real problem is that, you need to fix $\varepsilon$ first when $y\to x_0$, but this can not be done as what I just illustrated.

While in your second attempt, you send $\varepsilon$ to $0$, which forces $f(y)=f(x_0)$ by $$ f(x_0)-\epsilon<f(y)<M<f(y)+\varepsilon<f(x_0)+2\epsilon .$$

The reason why you can let $\varepsilon \to 0^+$ is that $x\to x_0,$ which promises that $|x-x_0|<\delta$ for any $\delta >0.$

You don't need $f(y)$ is fixed when $\varepsilon \to 0^+$.

So this is just the essential difference between your first attempt and the second one.

  • $\begingroup$ I am sorry that my answer may be a little bit tedious in advance. $\endgroup$
    – Sam Wong
    Nov 25 '20 at 16:07
  • $\begingroup$ Please see my edit :) $\endgroup$
    – Philipp
    Nov 25 '20 at 20:57
  • $\begingroup$ As @ancientmathematician mentions above, what you were trying to prove was nothing than $\lim_{x\to x_0}M(x)=M(x_0)=f(x_0),$ where $M(x):=\sup\{f(y)\mid y\in[x,x_0]\}$ which is a function of $x$. You can have $\lim_{x\to x_0}M(x)=M(x_0)$ only if $f(x)$ is continuous at $x_0$, which is the least condition you need. Because you only consider behavior of functions around $x_0$, you don't need $f(x)$ is continuous on the whole interval $[x,x_0].$ But the continuity of $f(x)$ at $x_0$ is crucial to your approach in the edit. $\endgroup$
    – Sam Wong
    Nov 26 '20 at 2:26
  • $\begingroup$ @Philipp You may want to sit down to think of the logic behind the $\epsilon - \delta$ language again. And before performing any limit action, make sure that you are clear about the dependence among all variables. And if you find my answer is useful, can you give me a upvote? Thanks. $\endgroup$
    – Sam Wong
    Nov 26 '20 at 2:32

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.