# Proof verification for "continuity of function $f\implies$ continuity of $m( x) =\min_{a\leq t\leq x} f( t)$".

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: $$$$0\leq m(c-)-m(c+)\leq m(x)-f(z)\le f(t)-f(z) \tag{2}$$$$

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.

• Comments are not for extended discussion; this conversation has been moved to chat. 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 $$x, but you seem to want to claim that $$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.

• 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 :)
– Koro
May 27 at 19:14
• 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. May 27 at 19:21
• Ah yes. That's what I wanted. Fixed the typo :)
– Koro
May 27 at 19:30
• 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 $c$. May 27 at 19:36
• 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?
– 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.