Is $1-\cos nt \leq n(1-\cos^nt)$?

Question

In an exercise about characteristic functions of probability measures, once is asked to show that $$ \begin{align} 1-\operatorname{Re}(\widehat{\mu}(nt))\leq n\Big(1-\big(\operatorname{Re}(\widehat{\mu}(t))\big)^n\Big)\tag{0}\label{zero} \end{align} $$ where $\widehat{\mu}(t)=\int e^{itx}\,\mu(dx)$, and $\mu$ is a Probability measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$.

Since $\Big(\int \cos(tx)\,\mu(dx)\Big)^n\leq \int\cos^n(tx)\,\mu(x)$, inequality $\eqref{zero}$ will follow immediately if the following inequality holds

$$ \begin{align} 1-\cos (n\alpha) \leq n\big(1-\cos^n(\alpha)\big),\quad\forall \alpha\in[0,2\pi]\tag{1}\label{one} \end{align} $$

• I tried induction (for n=1,2 we have equality, for n=3, $\eqref{one}$ holds) however, there does not seem the be a clear inductive proof.

• I tried to use mean value theorem, but again, the proof eludes me.

I have check numerically (just plotting the functions on the left and right hands sides) and it seems that the statement holds.

Does anybody have a hint or know of a proof (or counter example) for $\eqref{one}$?.

Answer 1

Separate Case: $\boldsymbol{n=0}$

Note that $$ 1-\cos(nt)\le n\!\left(1-\cos^n(t)\right)\tag1 $$ is trivially true for $n=0$.

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Prove a Stronger Statement for $\boldsymbol{n\ge1}$

To prove $(1)$ for $n\ge1$, we will prove the stronger statement $$ \left|\,n\cos^n(t)-\cos(nt)\,\right|\le n-1\tag2 $$ We have the two identities $$ (n+1)\cos^{n+1}(t)=n\cos^n(t)\cos(t)+\cos^n(t)\cos(t)\tag3 $$ and $$ \cos((n+1)t)=\cos(nt)\cos(t)-\sin(nt)\sin(t)\tag4 $$

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Induction on $\boldsymbol{n}$

Trivially, $(2)$ is true for $n=1$. Assume $(2)$ is true for some particular $n$, then $$ \begin{align} &\left|\,(n+1)\cos^{n+1}(t)-\cos((n+1)t)\,\right|\tag5\\[9pt] &=\left|\,\color{#C00}{\left(n\cos^n(t)-\cos(nt)\right)}\cos(t)+\color{#090}{\cos^n(t)}\cos(t)+\sin(nt)\sin(t)\,\right|\tag6\\[3pt] &\le\color{#C00}{(n-1)}|\!\cos(t)|+\color{#090}{\frac{n-1+|\!\cos(nt)|}n}|\!\cos(t)|+|\!\sin(nt)|\,|\!\sin(t)|\tag7\\ &=\frac{n^2-1+|\!\cos(nt)|}n|\!\cos(t)|+|\!\sin(nt)|\,|\!\sin(t)|\tag8\\[6pt] &\le n\tag9 \end{align} $$ Explanation:
$(6)$: subtract $(4)$ from $(3)$
$(7)$: triangle inequality and the assumption
$(8)$: simplify
$(9)$: Cauchy-Schwarz and the inequality
$\phantom{\text{(9):}}$ $\begin{align} \scriptsize\left(\frac{n^2-1+|\!\cos(nt)|}n\right)^2+\sin^2(nt) &\scriptsize\,=n^2-\frac{n^2-1}{n^2}(1-|\!\cos(nt)|)^2\\ &\scriptsize\,\le n^2 \end{align}$

Inequality $(9)$ shows that $(2)$ is true for $n+1$, which completes the induction.

Answer 2

The paper (here) has a a proof by induction of inequality (1) in the OP. They use the inequality $$ \begin{align} \Big|\frac{\sin(nx)}{\sin(x)}\Big|\leq n \tag{a}\label{a} \end{align} $$ which holds for all $n\in\mathbb{N}$ and $x\in\mathbb{R}$, to prove it.

Here is a sketch of the proof:

Set $f(x)=n\cos^nx -\cos(nx)$. Then, inequality (1) in the OP is equivalent to $$f(x)\leq n-1$$ for all $n\in\mathbb{N}$ and $x\in\mathbb{R}$. From

$$f'(x)=-n^2\cos^{n-1}(x)\,\sin(x) + n\sin(nx)$$

we have that all critical points $x$ of $f$ satisfy either $$ \begin{align} \sin(x)=0\tag{b}\label{b} \end{align} $$ or $$ \begin{align} \sin(x)\neq0,\quad\cos^{n-1}(x)=\frac{\sin(nx)}{\sin(x)}\tag{c}\label{c} \end{align} $$ If $x$ is a maximum point and satisfies $\eqref{b}$ then clearly $f(x)\leq n-1$ which is the the second inequality in the OP.

When $x$ is a maximum point and satisfies $\eqref{c}$, then $$ f(x)=\frac{\cos(x)\,\sin(nx)}{\sin x} - \cos(nx)=\frac{\sin(n-1)x}{\sin(x)}\leq n-1 $$

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The proof of $\eqref{a}$ can be found here