Let $\displaystyle f:[ a,b]\rightarrow \mathbb{R}$ be a continuous function, that is $\displaystyle f\in \mathcal{C}[ a,b]$. It is to be proven that $\displaystyle m\in \mathcal{C}[ a,b]$, where $\displaystyle m( x) =\min_{a\leq t\leq x} \ f( t)$.

I divide it into two parts:

First part: It can be shown that $\displaystyle m$ is monotonically decreasing.

Since $\displaystyle m$ is monotonic, its one sided limits exist. Idea is to show there is no discontinuity (jump discontinuity) at any $\displaystyle c\in [ a,b]$ and this is achieved by showing $\displaystyle m( c-) =m( c+)$.

Second part:

Fix any ${\displaystyle \delta >0}$, \begin{align} 0\leq m(c-)-m(c+)=&\inf \{m(x):a\leq x< c\}-\sup \{m(y):c< y\leq b\}\\ =&\inf \{m(x):c-\delta < x< c\}-\sup \{m(y):c< y< c+\delta \}\\ =&\inf \{m(x)-m(y):c-\delta < x< c< y< c+\delta \}\\ \Longrightarrow &0\le m(c-)-m(c+)\leq m(x)-m(y) \tag1 \end{align} Here in $(1)$, $x$ is any point in $(c-\delta,c)$ and $y$ is any point in $(c,c+\delta)$.

By continuity of $\displaystyle f$, minimum value is attained by $\displaystyle f$. In particular, $m(y)$ is attained by $f$ on $[a,y\rbrack$. Let's call the point $z$ where minimum value of $f$ on $[a,y\rbrack$ that is $m(y)$ is attained. There are two cases: Case 1): $z\le x$, Case 2): $y\ge z\gt x$.

Case 1): It follows that $m(y)=\min_{a\le t \le y} f(t)=f(z)$ for some $z\leq x$, whence it follows that $m(x)=f(z)$ (By definition of $m$).

It follows from ${\displaystyle (1)}$ that ${\displaystyle m(c-)=m(c+)}$ and we are done! Therefore, we have shown that from $z\le x$ it follows that $m(c-)=m(c+)$ thereby proving continuity at $c$.

Case 2): $y\ge z\gt x$
That is, $m(y)$ is attained by $f$ at $z\gt x$ that is ${\displaystyle m(y)=f(z)}$. Here the idea is to bring $t$ (this variable $t$ is introduced in $(2)$) and $z$ in some delta neighborhood of $c$ and this will help us apply uniform continuity later.

From ${\displaystyle (1)}$, we get: \begin{equation} 0\leq m(c-)-m(c+)\leq m(x)-f(z)\le f(t)-f(z) \tag{2} \end{equation}

Here ${\displaystyle t\in (c-\delta ,x)}$. Clearly right hand side on ${\displaystyle (2)}$ is positive.

By uniform continuity of ${\displaystyle f\ }$ on $\displaystyle [ a,b]$, for any ${\displaystyle \epsilon >0,}$ there exists a ${\displaystyle \delta _{1} >0}$, such that $ $for all ${\displaystyle p,q}$ in ${\displaystyle (c-\delta _{1} ,c+\delta _{1} )}$, we have ${\displaystyle |f(p)-f(q)|< \epsilon }$.

Now setting ${\displaystyle \delta =\delta _{1}}$, ${\displaystyle (2)}$ gives ${\displaystyle 0\leq m(c-)-m(c+)< \epsilon }$ and since ${\displaystyle \epsilon >0}$ is arbitrary, the result follows. If $\displaystyle c$ is end point $\displaystyle a$, then set $\displaystyle m( c-) =m( a)$ and the proof still works! Similarly for the other end.

Is my proof correct? Thanks.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    May 21 at 17:34

In Case (2), how do you know that $t\in (c-\delta,x)$? It seems like $t$ is defined by $f(t)=m(x)$ so all you can say about it a priori is that $t\in [a,x]$. Similarly all you know about $z$ is that $x<z\leq y$, but you seem to want to claim that $t,z\in (c-\delta_1,c+\delta_1)$ so that you can conclude $|f(t)-f(z)|<\epsilon$.

In Case 2, you can clean up the argument considerably by using the intermediate value theorem to conclude that $f(z')=m(x)$ for some $z'\in [x,z]$ (this follows because $f(x)>m(x)>f(z)$). By continuity of $f$, we have $|f(z')-f(z)|<\epsilon$ provided that $|x-y|<\delta_{\epsilon}$. But also $|f(z')-f(z)|=|m(x)-m(y)|$.

Also it seems like you are making things needlessly complicated by using one-sided limits, since you end up just bounding $m(x)-m(y)$ directly anyway.

  • 1
    $\begingroup$ Thanks a lot for your response. I understand that my proof got lengthier and that's why probably there's been less response to this post. I'm still learning :) I understand that there may be alternatives to it. Why I've posted this is to understand exactly what is wrong in my proof. Coming back to your query, why $t\in (c-\delta, x)$? I Please note that $m(x)$ is minimum value of $f(t)$ on $[a,x]$ so whatever $f(t)$ you take for any $t\in [a,x]$, we must have $m(x)\le f(t)$. That was the idea. If you feel more clarifications are required for the proof, please let me know. Thanks :) $\endgroup$
    – Koro
    May 27 at 19:14
  • 1
    $\begingroup$ OK I agree with that reasoning. In that case did you mean to write $m(x)-f(z)\leq f(t)-f(z)$ instead of $m(x)-f(z)=f(t)-f(z)$ in equation 2? The equals sign makes it look like you are defining a certain value of $t$ via an equation, and then asserting that this certain value lies in $(c-\delta,x)$, rather than writing an inequality that holds for all $t$ within the given range. $\endgroup$
    – Mike Hawk
    May 27 at 19:21
  • $\begingroup$ Ah yes. That's what I wanted. Fixed the typo :) $\endgroup$
    – Koro
    May 27 at 19:30
  • $\begingroup$ makes more sense now :). But I still don't see how you conclude that $m(c^-)-m(c^+)<\epsilon$, since you don't know how far $z$ is from $c$. $\endgroup$
    – Mike Hawk
    May 27 at 19:36
  • $\begingroup$ To see that how, please note that my $\delta\gt 0 $ is arbitrary :) and to achieve this $\lt \epsilon $, I use uniform continuity of $f$ to get a $\delta_1\gt 0$ and then setting $\delta=\delta_1$ concludes $\lt \epsilon $. Does it make more sense now? $\endgroup$
    – Koro
    May 27 at 19:45

This is just for fun.

For any $y \ge x$, we have $$ \min_{ x \le t \le y} f(t) =f(x)-\max_{ x \le t \le y} \big( f(x)-f(t) \big) \ge f(x)- \max_{x \le t \le y} |f(t)-f(x)| \ge m(x) -\max_{x \le t \le y} |f(t)-f(x)|$$ Thus, by definition $$m(x) \ge m(y)= \min\big( m(x), \min_{x \le t \le y} f(t) \big) \ge m(x) -\max_{x \le t \le y} |f(t)-f(x)|$$ Hence $$0 \le m(x)-m(y) \le \max_{x \le t \le y} |f(t)-f(x)| \le \max_{ u,v\in[x,y]} | f(u)-f(v)| $$ Which gives the conclusion.


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