Show that $\sin(f)$ is a continuous mapping in $\mathcal{B}(M)$

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.

• Boundedness of $f$ and $g$ ensure $\|f - g\|_\infty$ is finite.
– kobe
Commented Mar 17, 2022 at 12:36
• 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. Commented Mar 17, 2022 at 14:15
• @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 $\sin$. Commented Mar 18, 2022 at 12:42
• @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 $f\in B(M)$." Commented Mar 19, 2022 at 16:45
• @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. Commented Mar 20, 2022 at 19:30

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.

REFERENCE: