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Let $f \colon \mathbb R \to \mathbb R$. Suppose $f$ has a discontinuity at $a \in \mathbb R$. Now suppose also a continuous function $g \colon \mathbb R \to \mathbb R$ is given.

Is it true, then, that the function $g \circ f \colon \mathbb R \to \mathbb R$ has a discontinuity in $a$ as well?

Of course, this isn't true in general, since for instance $g$ could be a constant function, and then $g \circ f$ is obviously continuous in all points, so in particular in $a$. But, are there conditions we can give such that the statement would become true? E.g., perhaps we can assume that $g$ is non-constant in some neighborhood of $f(a)$, and this would suffice?

Thanks in advance.

herbhofsterd
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  • If $g$ is the identity function, what happens? – user24142 Jul 10 '23 at 03:35
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    I am not optimistic that you are going to come up with any particularly nice characterization, as there are all kinds of pathologies which might get smoothed out by a sufficiently compatible function. For example, take $f = \chi_{\mathbb{Q}} - \frac{1}{2}$ and $g = |\cdot |$. $f$ is highly discontinuous, but the composition with $g$ is the constant function $1/2$. – Xander Henderson Jul 10 '23 at 03:36
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    A simple class of counterexamples: If $f$ has a jump discontinuity and jumps from $y_1$ to $y_2$ at $a$, but $g(y_1)=g(y_2)$, then $g\circ f$ will be continuous at $a$. – Karl Jul 10 '23 at 03:39
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    I'm too tired to check the details, but my intuition tells me that if $f$ has an oscillatory discontinuity at $a$ and $g$ is continuous, then $g\circ f$ is continuous if and only if $g$ is constant on some sufficiently large interval around $f(a)$ (an interval large enough to capture the entire oscillation). – Xander Henderson Jul 10 '23 at 03:44
  • @XanderHenderson. Your intuition is correct. – DanielWainfleet Jul 10 '23 at 06:29
  • A jump discontinuity at $x= a$ is suppressed by $ x \ \to \ const. + g(x-a)$, s.t. $g(0)=0$ – Roland F Jul 10 '23 at 08:32
  • @XanderHenderson Nice 'counterexample'. Thanks. I think you meant that the composition is the constant function 1/2, but I'm sure it was a typo (and it doesn't make a difference for the argument anyways). – herbhofsterd Jul 10 '23 at 19:42
  • @herbhofsterd I don't know what you are talking about. :P (Typo corrected.) – Xander Henderson Jul 10 '23 at 20:08

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