不等式

Oscillation

Example:

\[ \begin{gather} b_n=c_n \, a_n=(-1)^{n+1}a_n \text, \, a_n \geq 0\\ \sum_{n=1}^{N} b_n \leq \sum_{n=1}^{N}\left| b_n \right| = \sum_{n=1}^N a_n \end{gather} \]

Use Euler's formula to rewrite

\[ (-1)^n=e^{2\pi in\frac{1}{2}}=\cos(2\pi n\frac{1}{2})+i\sin(2\pi n\frac{1}{2}) \]

Periodicity is 2. Moreover , Take periodicity k

\[ (e^{2\pi i\frac{1}{k}})^{n+k}= (e^{2\pi i\frac{1}{k}})^{n}\\ \]

Fourier series

More generally, for $ x$, consider \(e^{-i\omega n}\): Periodicity \(\frac{1}{x}\), and \(\omega = \frac{2\pi}{x}\)

\[ F(x):= \sum_{n=-\infty}^\infty a_n e^{-i\omega n}\text{ , } F_N(x):= \sum_{n=-\infty}^\infty a_n e^{-i\omega n} \]

Generate a new function on \(\left[-1,1\right]\) , known as Fourier analysis.

Since \(|e^{i\theta}|\leq 1\), therefore

introduction a norm

For a sequence \((a_n)_{n\in\mathbb{Z}}\in\mathbb{R}\), Let

\[ \left\|a_n \right\|_{\boldsymbol l^2}:= (\sum _{n=-\infty}^\infty |a_n|^p )^\frac{1}{p} \text{ for } p\in [1,\infty] \]

and

\[ \|a_n\| = \mathop{sup}\limits_{n\in\mathbb{Z}}|a_n| \]

We can measure the distance between two sequences e.g. for \(c_n=(-1)^{n+1}\) and \(d_n=(-1)^{n+1}(1+\frac{1}{n})\).

\[ \|c_n-d_n\|=(\sum_{n=-\infty}^{\infty}\frac{1}{n^2})^\frac{1}{2}=\frac{\pi}{\sqrt6}\leq2 \]

\(L^2\)-estimate: Crude

Recall the crude estimate :\(|F_N(x)| \leq \sum_{n=-N}^{n=N} a_n\) ：no oscillation

This yields

\[ \begin{aligned} \|F_N\|_{L^2}&=(\int_{-1}^1(F_N(x))^2d\,x)^\frac{1}{2}\\ &\leq(\int_{-1}^1a_n^2d\,x)^\frac{1}{2})\\ &=2^{\frac{1}{2}}\sum_{n=-N}^{N} a_n \end{aligned} \]

In particular, if \(a_n=1\) for all \(|n|\leq N\) then

\[ \|\sum_{n=-\infty}^\infty e^{-i\omega n}\| \leq 2^\frac{3}{2}N \]

\(L^2\) -size of \(F_N(x)\) is bounded by \(O(N)\)

Theorem 3 (Parseval's identity)

\[ \|\sum_{n=-\infty}^\infty e^{-i\omega n}\| \leq 2N^\frac{3}{2}\ll2^\frac{3}{2}N \]

Multiplying out the square:

\[ \begin{aligned} \|\sum_{n=-N}^N e^{-i\omega n}\| &=(\int_{-1}^1(\sum_{n=-N}^N e^{-i\omega n}) ^2d\,x)^\frac{1}{2}\\ &=\int_{-1}^1 \sum_{n=-N}^N e^{-i\omega n}\sum_{m=-N}^N e^{-i\omega n}d\,x\\ &= \sum_{n=-N}^N\sum_{m=-N}^N \int_{-1}^1 e^{-i\omega (n-m)}d\,x \end{aligned} \]

For the diagonal case (m=n)

\[ \int_{-1}^1 e^{-i\omega (n-m)}d\,x =2 \]

For the non-diagonal case (\(n\neq m\))

\[ \int_{-1}^1 e^{-i\omega (n-m)}d\,x =-\frac{1}{2\pi i (n-m)}(e^{-2\pi i (n-m)}-e^{2\pi i (n-m)})=0 \]

Same argument shows more generally

\[ \begin{gather} \text{For any }a_n \;s.t.\|a_n\|_{l^2}=(\sum_{n=-\infty}^\infty |a_n|^2)^{\frac{1}{2}}<\infty.\\ \|\sum_{n=-\infty}^\infty e^{-i\omega n}\| _{l^2}=2^{\frac{1}{2}}\|a_n\|_{l^2} \end{gather} \]