composition function with dense graph

Question

Let $f\colon \Bbb R\to \Bbb R$ be a function with dense graph .Let $g\colon \Bbb R \to \Bbb R$ be a function( Not constant).

Does $f\circ g$ have a dense graph? if not, How about if $g$ is a homeomorphism? my guess is not but I could not have an example?

@HennoBrandsma, I forgot to add that not constant. How about in general. ? — 00GB, Mar 15 '21 at 15:03
Examples are nasty as such functions are hard to write down. There are plenty of them, you can construct them by transfinite recursion, but not many explicit ones. — Henno Brandsma, Mar 15 '21 at 15:06
@HennoBrandsma, you meant we construct $g$ such that $f\circ g$ has dense graph.. Right? OKay , For $g$ is a homeomorphism , what we can say about the graph of $f\circ g$ — 00GB, Mar 15 '21 at 15:10
Maybe it's true that if $g$ is a homeomorphism, unequal to the identity, there exists $f$ with dense graph such that $f \circ g$ does not have a dense graph. — Henno Brandsma, Mar 15 '21 at 15:13
This is a variant of https://math.stackexchange.com/q/4061578. You should consider to close one of them. — Paul Frost, Mar 16 '21 at 00:29

hartkp · Answer 1 · 2021-03-16T07:12:16.363

0

Here are some examples to give an idea of what is possible.

if $f$ is a bijection and $g$ is its inverse then $f\circ g$ is the identity, so its graph is not dense
if $g$ is a homeomorphism then the graph of $f\circ g$ is the image of the graph of $f$ under the product map $g^{-1}\times\mathrm{Id}$, which is a homeomorphism, so $f\circ g$ is dense.
for many `common' continuous functions the result is again a dense graph, e.g., $g(x)=|x|$ takes the part of the graph of $f$ over $(0,\infty)$ and reflects it with respect to the $y$-axis, or $\sin x$ takes the part over $[-1,1]$ and stretches it to $[-\frac\pi2,\frac\pi2]$ and then alternately shifts it or its mirror image over the $x$-axis
if $g$ is constant on an interval then $f\circ g$ is constant over this interval and the graph is not dense

and so on

edited Mar 16 '21 at 07:12

answered Mar 15 '21 at 19:39

hartkp

1

hartkp, Thank you, but this does not address the OP since your $f$ does not have a dense graph. – 00GB Mar 15 '21 at 20:27
I'm working under the standing assumption that $f$ has a dense graph. In the first point it is, in addition, a bijection. In the other points it is only assumed to have a dense graph. It is $g$ that is being varied here. In cases 1 and 4 the resulting graph is not dense, in cases 2 and 3 it is. – hartkp Mar 16 '21 at 06:52

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