Let $f(x)$ be a continuous function on $\Bbb X \subset \Bbb R$. Let $a, b \in\Bbb R$ and $a < b$. Prove that the following function is also continuous on $\Bbb X$: $$ f(a, b, x) = \begin{cases} f(x),\ \text{if $a \le f(x) \le b$} \\ a,\ \text{if $f(x) < a$}\\ b,\ \text{if $f(x) > b$} \end{cases} $$

Intuitively the function above is defined in a way to "cut" the upper and lower parts of $f(x)$. So my idea was to define a new function in terms of $\min$ and $\max$ functions. I've started by defining a couple of helper functions. Let: $$ g(x) = \min\{b; f(x)\}\\ h(x) = \max\{a; f(x)\} $$

Here is a visualization in desmos I've been playing with. Using $g(x)$ and $h(x)$ we might redefine $f(x)$ as: $$ f(x) = \max\left\{\min\{g(x); h(x)\}; a\right\} $$

Consider min and max functions in the following form: $$ \begin{align*} \max\{x, y\} = {1\over 2}(x + y) + {1\over 2}|x-y|\tag1\\ \min\{x, y\} = {1\over 2}(x + y) - {1\over 2}|x-y|\tag2 \end{align*} $$

Using the above definition it follows that both min and max are continuous in case $x$ and $y$ are continuous by composition of continuous functions. Since $c_1(x) = b$ is a constant function and $c_2(x) = a$ is also a constant function, then they are both continuous. $f(x)$ is also continuous by initial conditions.

Using these facts we obtain: $g(x)$ and $h(x)$ are continuous by composition of continuous functions is continuous.

At this point, I'm not sure whether I need to express $f(x)$ in terms of $(1)$ and $(2)$. I mean we could rewrite $f(x)$ as: $$ f(x) = {1\over 2}(\min\{g(x); h(x)\} + a) + {1\over 2}|\min\{g(x); h(x)\} - a| = \cdots $$

And we could continue the same way until we get a huge expression in terms of $(1)$ and $(2)$ without mentioning min and max. So eventually we would again obtain a composition of continuous functions which is also continuous. But that is redundant, isn's that?

I'm interested whether the approach above is even correct? If not, what would be a way to show $f(a,b,x)$ is continuous? Also, could we find a more simple approach?

Thank you!

  • $\begingroup$ I like this approach. I think it is the best one, in fact. $\endgroup$ – Kavi Rama Murthy Jul 26 '19 at 10:18

(Remark: I like the proof as posted, especially the elegant construction of min and max. The following is merely an alternative.)

For clarity define $g(x) \equiv f(a,b,x)$. Recall that $g(x)$ is continuous if and only if $g^{-1}(S)$ is open whenever $S$ is open. So let $S$ be any open subset of $[a,b]$ (the codomain of $g$). Now $g^{-1}(S) = f^{-1}(S)$ by construction. Moreover $f$ is continuous so $f^{-1}(S)$ is open. Result follows.


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