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Let $f, g:\mathbb R\to\mathbb R$ such that $g\in L^\infty(\mathbb R)$ and $f\in C(\mathbb R)$ (i.e. $f$ is a continuous function over $\mathbb R$).

According to me $$f(g):x\in (-2, 2)\mapsto f(g(x))$$ satisfies $f(g)\in L^\infty ((-2, 2))$ since $f$ is continuous and $g$ bounded (the (-2,2) can be replaced with any other symmetric bounded interval).

On the other hand, I am not sure if I can say that $f(g)\in L^\infty(\mathbb R)$.

Anyone could please help me in understanding this? Thank you in advance.

C. Bishop
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  • Yes, it is bounded, see e.g. https://math.stackexchange.com/questions/3452923/composition-of-continuous-and-bounded-function-is-bounded – LSK21 Nov 21 '23 at 10:27
  • According to the question you mentioned, it seems that it is bounded only over bounded set, not over the entire $\mathbb R$. – C. Bishop Nov 21 '23 at 12:05
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    See in particular the second answer by Kavi Rama Murthy at the link. The point is that in your case $f$ is continuous on all of $\Bbb R$, and therefore $f$ maps bounded sets in $\Bbb R$ to bounded sets. The image of the composition $f \circ g$ is the image of $\operatorname{im} g$ under $f$, which is the image of a bounded set under $f$, which must be bounded. – Izaak van Dongen Nov 21 '23 at 15:24

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