# What is the background to inequality $|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$?

$$\alpha=(x_1,x_2,\dotsc,x_n)$$, $$\beta=(y_1,y_2,\dotsc,y_n)\in\Bbb R^n$$, $$\alpha \cdot\beta=x_1y_1+x_2y_2+\dotsb+x_ny_n$$.

$$|\alpha|=\sqrt{x_1^2+x_2^2+\dotsb+x_n^2}$$, $$|\beta|=\sqrt{y_1^2+y_2^2+\dotsb+y_n^2},$$

Show that : there exists a constant $$L\gt 0$$ such that $$|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$$ for all $$\alpha,\beta$$

I don't know where to start

The question produces background . What is the background?

Well for $$t\in\mathbb{R}$$ $$|e^{it}-1|^2=(cos(t)-1)^2+\sin(t)^2=2-2\cos(t)<2|t|^2$$ so $$|e^{it}-1|\leq\sqrt{2}|t|$$ for all $$t$$. Now for $$t=\alpha\cdot\beta$$ we have that $$|e^{i\alpha\beta}-1|\leq\sqrt{2}|\alpha\cdot\beta|$$ and using the famous Cauhcy-Schwarz inequality $$|\alpha\cdot\beta|\leq|\alpha|\cdot|\beta|$$ we get the desired estimate.

• You can get a tighter bound by using MVT or $|\sin x|\leq |x|$ from your work by using the cosine double angle identity. Commented Feb 7, 2021 at 19:30
• @NinadMunshi Oh yeah, sure! I just used a quick "overkill", of course there are better estimates, I think that the optimal value must be 1, if I am not mistaken. Commented Feb 7, 2021 at 20:08

$$\textbf{Hint}$$: First prove that

$$|e^{ix}-e^{iy}|\leq L|x-y|$$

for $$x,y\in\Bbb{R}$$

$$e^{\,i\, \alpha \cdot \beta}$$ is always less than $$2$$, whatever is $$\alpha \cdot \beta=|\alpha| \, |\beta|\, \cos \theta$$.
So we have to consider the case in which $$\alpha \cdot \beta$$ is small.

Since the arc $$\alpha \cdot \beta$$ is comprised between $$\pm |\alpha|\,|\beta|$$, then the chord $$|e^{\,i\, \alpha \cdot \beta}|$$ is not greater that the chord $$|e^{|\alpha| \, |\beta|}|$$, which in turn is less then the arc $$|\alpha| \, |\beta|$$, which remains true also when the arc equals $$\pi$$.

Thus $$L=1$$