On uniformly continuous, bounded, hermitian functions whose product is continuously differentiable Let $f,g: \mathbb R \to \mathbb C$ be uniformly continuous, such that $f(0)=g(0)=1$, $|f(x)|\le 1, |g(x)|\le 1,\forall x\in \mathbb R$, $f(-t)=\overline {f(t)}, g(-t)=\overline {g(t)}, \forall t \in \mathbb R$ and $f.g$ is continuously differentiable on $\mathbb R$. 

Then is it true that $f,g$ are differentiable at $0$ ? If this is not true in general, what if we also assume that $f,g$ are real-valued ?
Define $r(x)=\frac1{|x|+1}$ and take $f(x)=r(x)+i\sigma(x)\sqrt{1-(r(x))^2}$ and $g(x)=\overline{f(x)}$. Here $\sigma(x)$ is the sign of $x$. Notice that $r$ is not differentiable at $0$, so neither are $f$ and $g$. However $|f(x)|=1$, so $fg=1$, so $fg$ is continuously differentiable. All the other properties are easier to check.

Now, to prove that the conjecture does hold for $f,g:\mathbb{R}\to\mathbb{R}$. If $f$ is not differentiable at $0$, then there exists $\varepsilon&gt;0$ such that for all $\delta&gt;0$ there exists $x\in(-\delta,\delta)$ with $1-f(x)&gt;\varepsilon|x|$. Since $|g(x)|\leq1$, we then also find $1-(fg)(x)&gt;\varepsilon|x|$ (for $x$ small enough such that $f(x)&gt;0$). Hence $fg$ is not differentiable at $0$.
Take $f(x)=(1+|x|)^{-0.5}$ then $f$ is even and hence $f(-x)=\overline{f(x)}$ since it is a real valued. It is easy to see that $|f(x)|\leq 1$ but $f$ is not differentiable at $0.$ Since its derivative is bounded for $|x|\geq 1$ the function is uniformly continuous.