Question: I have a map $f(t,g,h)$ where $f:[0,1]\times C^1 \times C^1 \to \mathbb{R}.$

I want to define $$F(t,g,h) = \frac{d}{dt}f(t,g,h)$$
where $g$ and $h$ have no $t$-dependence in them. So $g(x) = t^2x$ would not be admissible if you want to calculate what $F$ is. How do I write this properly? Is it correct to write instead

> 
> Define $F$ by $F(t,\cdot,\cdot) = \frac{d}{dt}f(t,\cdot,\cdot)$.
> 
> 
> 

But there is some ambiguity in the arguments. What is the best way to write it?

Answer: 
Without words... Well, nearly without words...

![enter image description here](https://i.stack.imgur.com/la68x.png)
The sine function is an odd function, so it suffices to prove the inequality for nonnegative $x$.

For $0\le x\le 1$, rewrite $\sin x$ in two different ways:
\begin{align\*}
\sin x &=\sum\_{n=0}^\infty\left(\frac{x^{4n+1}}{(4n+1)!} - \frac{x^{4n+3}}{(4n+3)!}\right)\tag{1}\\
&=x - \sum\_{n=0}^\infty\left(\frac{x^{4n+3}}{(4n+3)!} - \frac{x^{4n+5}}{(4n+5)!}\right).\tag{2}
\end{align\*}
Since each bracket term in the two infinite series is nonnegative, $(1)$ implies that $\sin x\ge0$ and $(2)$ implies that $\sin x\le x$. Therefore $|\sin x|\le |x|$.

For $x>1$, note that $\sin x = |\operatorname{imag}(e^{ix})| \le |e^{ix}|$. If we can show that $|e^{ix}|\le1$, we are done. One way to prove this is to show that $\overline{e^{ix}}=e^{-ix}$ (easy) and $e^{ix}e^{-ix}=1$ (theorems about multiplication of absolutely convergent power series, i.e. convergence of Cauchy product, is needed here).