To show that $\frac{f(x-t)-f(x)}{\sin(t/2)}$ is bounded on $[-\pi,\pi]$

Question

Given that $f$ is Lipschitz, I am trying to show that $\frac{f(x-t)-f(x)}{\sin(t/2)}$ is bounded on [$-\pi,\pi$]. But I am not getting how I will bound the denominator. What I did is as follows:

$\left|\frac{f(x-t)-f(x)}{\sin(t/2)}\right| \leq C \left\rvert\frac{t}{\sin(t/2)} \right\rvert$. I saw the graph of function, it is clearly bounded but I am failing to give mathematical reason. Any hint.Thanks.

Answer 1

You have done it almost completely, observe that now you only need to argue that $\frac{t}{\sin\frac{t}{2}}$ is bounded on $[-π,π]$, which is same as saying $\frac{x}{\sin x}$ is bounded on $[-\frac{π}{2},\frac{π}{2}]$. Now, that is a continuous function on compact interval, hence bounded.

Answer 2

I would use the Mean Value Theorem: $$f(b)-f(a)=f'(\xi)(b-a)$$