Exercise 11 Chapter 3 From Stein's Fourier Analysis

Question

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I have a proof for A and question for B

For part a), basically we want to apply Parseval's identity to $f$ and compare to its derivative. T-periodic, continuous and piecewise $C^1$ imply that $f$ and $f'$ are square integrable. So, we can apply Parseval's identity to both of them. Let $$f\sim \sum_{}{} a_ne^{2n\pi it/T } (\text{ so, } f'\sim \sum_{}^{} a_n2n\pi i/Te^{2n\pi it/T }).$$

Then, Parseval's identity; $\|f\|^2=\sum_{n=-\infty}^{\infty} |a_n|^2 $, implies

$$\frac{4\pi^2}{T^2}\Big(\frac{1}{T}\int_{0}^{T}|f|^2dt \Big)=\frac{4\pi^2}{T^2} \Big(\sum_{n=-\infty}^{\infty} |a_n|^2 \Big)\leq \sum_{n=-\infty}^{\infty} \Big|\frac{a_n2n\pi}{T}\Big|^2=\|f'\|^2=\frac{1}{T}\int_{0}^{T}|f'|^2dt, \text{ with equality iff $n=1.$}$$

Remark: The condition $\int_{0}^{T}f=0$ implies $a_0=0$, so, the above inequality is valid for all $n$.

For part b), I try to use the result from part A. From Cauchy-Schwarz Inequality; $|(f,g)|\leq \|f\| \cdot \|g\|,$

$$\Big|\frac{1}{T}\int_{0}^{T}\overline{f}g \ dt \Big|^2\leq \frac{1}{T^2}\int_{0}^{T}|f|^2dt\int_{0}^{T}|g|^2dt. $$

Thus, $$\Big|\int_{0}^{T}\overline{f}g \ dt \Big|^2\leq \int_{0}^{T}|f|^2dt\int_{0}^{T}|g|^2dt.$$

Here, it seems the result is intermediate if $g$ has the same property that $f$ had in part A. But, this $g$ is just T-periodic and $C^1$. Can we still reach the same result?

Answer 1

Use Plancherel to write the inner product in Fourier space, and then use Cauchy-Schwarz.

Answer 2

Here is my attempt: Writing the integrand in the Fourier series one has (in the following the integrals are always from $0$ to $T$)

$$ \begin{aligned} \int\overline{f} g &= \int (\sum_n \overline{a_n} e^{-i\frac{2\pi}{T}nt}) (\sum_m b_m e^{i\frac{2\pi}{T}mt}) \\ &= \sum_{m,n}\int \overline{a_n}b_n e^{i\frac{2\pi}{T}(m-n)t}\\ &=\sum_{m=n}\int \overline{a_n}b_n e^{i\frac{2\pi}{T}(m-n)t}\\ &=T\sum_{n}\overline{a_n}b_n \,. \end{aligned} $$

Thus

$$ |\int\overline{f} g|^2 = T^2 |\sum_n \overline{a_n}b_n|^2 \leq T^2 \sum_n|a_n|^2 \sum_n |b_n|^2 = T^2 \sum_n|a_n|^2 \sum_{n\neq 0} |b_n|^2, $$

where $\sum_n|a_n|^2 \sum_{n} |b_n|^2 = \sum_n|a_n|^2 \sum_{n\neq 0} |b_n|^2$ since $a_0 = 0$ (the result in part (a)).

Then from the first part of the problem we know

$$ \sum_{n\neq 0} |b_n|^2 \leq \sum_{n\neq0}n^2|b_n|^2 = \frac{T^2}{4\pi^2}(\frac{1}{T}\int |g'|^2), $$

and by Parseval's equality

$$ \sum_n |a_n|^2 = \frac{1}{T}\int|f|^2. $$

Putting these together we have

$$ |\int \overline{f}g|^2 = T^2 \sum_n|a_n|^2 \sum_{n\neq 0} |b_n|^2 \leq T^2 (\frac{1}{T}\int|f|^2) (\frac{T^2}{4\pi^2}\frac{1}{T}\int |g'|^2). $$

That is

$$ |\int \overline{f}g|^2 \leq \frac{T^2}{4\pi^2} \int|f|^2 \int |g'|^2. $$