Show that $\sin:\mathcal{B}(M)\to \mathcal{B}(M)$ is a continuous mapping, where $\mathcal{B}(M)$ is the space of bounded real valued functions equipped with the $\Vert\cdot\Vert_{\infty}$ sup norm.

My approach:

We already know that $h:\mathbb{R}\to\mathbb{R}$ where $h:=\sin(x)$ is a uniformly continuous function. So for an arbitrary $\epsilon>0$ we find a $\delta>0$ such that for all $x,y\in\mathbb{R}$ we have $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$.

Now we take two $f,g\in\mathcal{B}(M)$ such that $\Vert f-g\Vert_{\infty}<\delta$. This implies for all $x\in M$ that $|f(x)-g(x)|<\delta\implies |\sin(f(x))-\sin(g(x))|<\epsilon\implies \Vert \sin(f)-\sin(g)\Vert_{\infty}<\epsilon$.

Is this correct? I am a bit skeptical because it seems that we don't need the boundedness.

  • 1
    $\begingroup$ Boundedness of $f$ and $g$ ensure $\|f - g\|_\infty$ is finite. $\endgroup$
    – kobe
    Commented Mar 17, 2022 at 12:36
  • $\begingroup$ You need also (to be a bit pedantic) to claim that if $f\in B(M)$ then $\sin(f)$ is also a member of that space. That will use boundedness. $\endgroup$ Commented Mar 17, 2022 at 14:15
  • $\begingroup$ @B.S.Thomson, Why? If $\sin$ wasn't a bounded function, then this would not invalidate the result $|f(x)-g(x)|<\delta\implies |\sin(f(x))-\sin(g(x))|<\epsilon\implies \Vert \sin(f)-\sin(g)\Vert_{\infty}<\epsilon$, I only used/needed the uniform continuity of $\sin$. $\endgroup$
    – Philipp
    Commented Mar 18, 2022 at 12:42
  • $\begingroup$ @Philipp I posted an "answer" to illustrate the issue. Your answer is fine if you simply add something like: "It is obvious that $\sin(f)\in B(M)$ whenever $f\in B(M)$." $\endgroup$ Commented Mar 19, 2022 at 16:45
  • 1
    $\begingroup$ @Philipp You can "assume" in this case since it is true. In general, however, any statement about a mapping between two spaces would normally require you to check that it is in fact a mapping between those spaces. $\endgroup$ Commented Mar 20, 2022 at 19:30

1 Answer 1


The OP and I are debating a subtle point, too subtle for the tiny space of comments.

His solution of the problem is fine, but missing a necessary remark at least. This "answer" just addresses that point.

  1. For any bounded function $f:[0,1]\to \mathbb R$ define $\|f\|_\infty=\sup\{|f(x)|:x\in [0,1]\}$.

  2. Define $D[0,1]$ to be the linear space of all bounded derivatives $f:[0,1]\to \mathbb R$ equipped with the norm $\|f\|_\infty$. This is a normed linear space. In fact it is a much-studied Banach space. See reference [1].

  3. Let $F:\mathbb R\to \mathbb R$ be a uniformly continuous function and write $T_F$ for the mapping $f\to F\circ f$, i..e., this takes any real-valued function $f(x)$ and yields the function $F(f(x))$.

Prove the false statement that $T_F$ is a continuous map from the space $D[0,1]$ to itself.

Copying from the OP:

We already know that $F:R→R$ is a uniformly continuous function. So for an arbitrary $ϵ>0$ we find a $δ>0$ such that for all $s,t∈R$ we have $|s-t|<δ⟹|F(s)−F(t)|<ϵ.$

Now we take two $f,g∈D[0,1]$ such that $ ∥f−g∥_∞<δ$. This implies for all $x∈[0,1]$ that $|f(x)−g(x)|<δ⟹|F(f(x))−F(g(x))|<ϵ⟹∥T_F(f)−T_F(g)∥_∞<ϵ$. Thus $T_F$ is a continuous map.

Spoiler: One can give an example of a function $f\in D[0,1]$ such that $f^2$ is not in $D[0,1]$. In fact given any continuous, increasing function $F$ that is not linear there exists a bounded derivative $f$ such that $F\circ f$ is not a derivative.


[1] https://www.jstor.org/stable/44151124


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