$f(x)$ is continuous at $x=\alpha$ ,$g(x)$ is discontinuous at $x=a$ but $g(f(x))$ is continuous at $x=\alpha$

Question

Suppose $f,g:\mathbb{R} \rightarrow \mathbb{R}$ are such that $f(x)$ is continuous at $x=\alpha$ and $f(\alpha)=a$ and $g(x)$ is discontinuous at $x=a$, but $g\big(f(x)\big)$ is continuous at $x=\alpha$. Also, $f(x),g(x)$ are non-constant functions. Then, can it be said that $x=\alpha$ is an extremum of $f$ and $x=a$ is an extremum of $g$?

I have tried to construct examples of functions, but never could figure out a rigorous proof For example take the function $f(x)=x^2$ which is continuous at $0$, $g(x)=[x]$ which is discontinuous at $0$, but $g\big(f(x)\big)$ is continuous at $0$.

Answer 1

Here are some counterexamples, using the terminology of OP, i.e. $f$ continuous in $\alpha$, $g$ not continuous in $a=f(\alpha)$ and $g \circ f$ continuous in $\alpha$, always with $\alpha = 0$ and $a = 0$.

In this example $\alpha$ is an extremum for $f$ but $a$ is not an extremum for $g$: \begin{eqnarray} f(x) &=& x^2\\ g(x) &=& \begin{cases} x+1 & (x \geq 0)\\ x-1 & (x < 0). \end{cases} \end{eqnarray}

In the following example $\alpha$ is not an extremum for $f$, but $a$ is an extremum for $g$: \begin{eqnarray} f(x) &=& \begin{cases} x & (x\ \ \mbox{rational})\\ 0 & (x\ \ \mbox{irrational}) \end{cases} \\ g(x) &=& \begin{cases} 1 & (x\ \ \mbox{rational})\\ 0 & (x\ \ \mbox{irrational}). \end{cases} \end{eqnarray}

Finally an example where neither $\alpha$ nor $a$ are extrema of $f$ and $g$ respectively: \begin{eqnarray} f(x) &=& \begin{cases} x & (x\ \ \mbox{rational})\\ 0 & (x\ \ \mbox{irrational}) \end{cases} \\ g(x) &=& \begin{cases} 1 & (x\ \ \mbox{rational})\\ x & (x\ \ \mbox{irrational}). \end{cases} \end{eqnarray} Concerning the last two examples, note that $f(\mathbb R) \subseteq \mathbb Q$, so that $g \circ f$ is constant.

Answer 2

Here is a possible variation of OP. I don't think it is of much interest, and worked it out just to do a little practice myself. But you might want to have a look and see if it works.

Consider two functions $f,g:\mathbb R \rightarrow \mathbb R$, such that $f$ is continuous in $\alpha$ and not constant in any neighborhood of $\alpha$. Let then $f(\alpha) = a$ and suppose that \begin{equation} \lim_{x\rightarrow a^+} g(x) = \ell_+ \tag{1}\label{piu} \end{equation} and \begin{equation} \lim_{x\rightarrow a^-} g(x) = \ell_- \tag{2}\label{meno} \end{equation} exist and they are different, so that $g(x)$ is not continuous in $a$. If $(g\circ f)(x)=g(f(x))$ is continuous in $\alpha$, then $\alpha$ is an extremum of $f$.

Proof. Suppose that $\alpha$ is not an extremum of $f$. Since $f$ is not constant in any neighboorhood of $\alpha$, then in the neighborhood \begin{equation} I_{2n} = \left(\alpha-\frac{1}{2n}, \alpha+\frac{1}{2n}\right),\ n\in \mathbb Z^+ \end{equation} there must be a value $\alpha_{2n-1}$ for which $f(\alpha_{2n-1})< a$ and a value $\alpha_{2n}$ for which $f(\alpha_{2n})> a$.

The sequence $(\alpha_n)$ converges to $\alpha$. Thus, by continuity of $f$, $(f(\alpha_n))$ converges to $f(\alpha) = a$. However, the sequence $(g(f(\alpha_n)))$ does not converge, since the subsequence $(g(f(\alpha_{2n})))$ converges to $\ell_+$, by \eqref{piu}, and the subsequence $(g(f(\alpha_{2n-1})))$ converges to $\ell_-$, by \eqref{meno}. This contradicts continuity of $g\circ f$ in $\alpha$.