• Title: Bessel’s Inequality and Parseval’s Identity

• Series: Fourier Transform

• Chapter: Fourier Series

• YouTube-Title: Fourier Transform 8 | Bessel’s Inequality and Parseval’s Identity

• Bright video: https://youtu.be/nRuLT4l8DGU

• Dark video: https://youtu.be/Na_O7SZQKkA

• Subtitle on GitHub: ft08_sub_eng.srt missing

• Timestamps (n/a)
• Subtitle in English (n/a)
• Quiz Content

Q1: Consider $V = \mathbb{R}^4$ with the standard inner product and the ONS given by $(e_1, e_2, e_3)$, where $e_1 = (1,0,0,0)^T$, $e_2 = (0,1,0,0)^T$, and $e_3 = (0,0,1,0)^T$. How does Bessel’s inequality look like in this case for $x = (x_1, x_2, x_3 , x_4)^T \in V$.

A1: $$\sum_{k = 1}^4 | x_k | \leq | x |^2$$

A2: $$\sum_{k = 1}^3 | x_k |^2 \leq | x |$$

A3: $$\sum_{k = 1}^3 | x_k |^2 \leq \sum_{k = 1}^4 | x_k |^2$$

A4: $$\sqrt{\sum_{k = 1}^2 | x_k |^2} \leq | x |^2$$

Q2: Consider a finite-dimensional vector space $V$ with the standard inner product and an ONB $(e_1, \ldots, e_n)$. Is Parseval’s identity satisfied in this case?

A1: Yes, we always have $$\sum_{k = 1}^n |\langle e_k, x \rangle|^2 = | x |^2$$ for all $x \in V$.

A2: No, only in special cases. We only have $$\sum_{k = 1}^n |\langle e_k, x \rangle|^2 \leq | x |^2$$ for all $x \in V$ in general.

A3: No, it is never fulfilled.

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