Continuous function with parameter

Question

Let $a>0$ be a real number, $f\colon \mathbb R \to \mathbb R$ a continuous function and $g\colon \mathbb R \to \mathbb R$ the function given by $$g(x)=\begin{cases} -a & if & f(x)<-a\\ f(x) & if & |f(x)|\le a\\ a & if & f(x)>a\\ \end{cases}.$$

Prove that $g$ is also a continuous function.

I managed to solve it with new function I assembled but maybe you will have better ideas.

thanks.

Answer 1

What's your try?

I would go to prove for arbitrary $x_0 \in \mathbb R$ that $g$ is continuous in $x_0$, that ir: $\lim_{x\to x_0} g(x) = g(x_0)$.

But what the value of $g(x_0)$ is depends on $f(x_0)$. In any case, the continuity of $f$ will be central.

If $|f(x_0)|<a$ (note that I said strictly smaller), then $g(x_0)=f(x_0)$, but what about the limit? Well, since $|f(x_0)|<a$, there has to be an open interval around $x_0$ for which this condition still holds, more precisely: $$ \exists \delta>0 \colon \qquad |f(x)|<a \quad \forall x\in (x_0-\delta,x_0+\delta).^*$$

But then $$ \exists \delta>0 \colon \qquad g(x)=f(x) \quad \forall x\in (x_0-\delta,x_0+\delta).$$ and so $g$ is continuous as $f$ is (they are the same function in an open interval, so their limits coincide in every points of it).

To complete the proof you have to say that if $f(x_0)>a$ or $f(x_0)<a$, there is an open interval on which that inequality holds and so $g(x)=\pm a$ on that open interval and so the limit is that same constant. And finally, you should work over the cases $|f(x)|=a$ combining ideas.

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$^*$ This would be a more generic use of the so-called sign conservation property, which states that if $f$ is continuous at $x_0$ and $f(x_0)>0$, then $f(x)>0$ over an open interval containing $x_0$ too. The same is true for $f(x_0)<0$, of course, but is obviously false for $f(x_0)=0$. But the $0$ is arbitrary, and we could state the same property for any $a\neq 0$.

Answer 2

I will define a new function $h$ like that:

$h\colon \mathbb R \to \mathbb R$ $$h(x)=\begin{cases} -a & if & x<-a\\ x & if & |x|\le a\\ a & if & x>a\\ \end{cases}.$$

Abviously, that function is continuous function (we can calculate that by limits for both sides of $a$ and $-a$)

now, I'll assemble $f(x)$ on $h(x)$ we will have excatly $g(x)$.

Assemble of 2 continuous functions will be continuous.