---
title: "cornucopia of analysis"
type: concept
source: https://www.jemoka.com/posts/kbhcornucopia_of_analysis/
confidence: high
status: active
---

Pythagorean Theorem \begin{equation} \|u + v\|^{2} = \|u \|^{2} + \|v\|^{2} \end{equation}
if \(v\) and \(u\) are orthogonal vectors.
Proof:
An Useful Orthogonal Decomposition Suppose we have a vector \(u\), and another \(v\), both belonging to \(V\). We can decompose \(u\) as a sum of two vectors given a choice of \(v\): one a scalar multiple of \(v\), and another orthogonal to \(v\).
That is: we can write \(u = cv + w\), where \(c \in \mathbb{F}\) and \(w \in V\), such that \(\langle w,v \rangle = 0\).
Here&rsquo;s how:
For nonzero \(v\)
\begin{equation} c = \frac{\langle u,v \rangle}{\|v\|^{2}} \end{equation}
and
\begin{equation} w = (u - cv) \end{equation}
We can show \(\langle w,v \rangle=0\) as follows:
\begin{align} \langle (u-cv), v \rangle &amp;= \langle u,v \rangle - \langle cv, v \rangle \\ &amp;= \langle u,v \rangle - c \langle v,v \rangle \\ &amp;= \langle u,v \rangle - \frac{\langle u,v \rangle}{\|v\|^{2}} \langle v,v \rangle \\ &amp;= \langle u,v \rangle - \frac{\langle u,v \rangle}{\|v\|^{2}} \|v\|^{2} \\ &amp;= 0 \end{align}
Cauchy-Schwartz Inequality \begin{equation} \qty(\sum_{k=1}^{n} a_{k} b_{k})^{2} \leq \qty(\sum_{k=1}^{n} a_{k}^{2})\qty(\sum_{k=1}^{n} b_{k}^{2}) \end{equation}
or more generally
\begin{equation} | \langle u,v \rangle | \leq \|u\| \|v\| \end{equation}
and the expression is an equality of each vector \(u,v\) is the scalar multiple of the other.
Proof:
Pick some set of \(v\) and \(u\) and write out the orthogonal decomposition we had outlined above:
\begin{equation} u = cv + w \end{equation}
Now, recall \(c = \frac{\langle u,v \rangle}{\|v\|^{2}}\). We now apply Pythagorean Theorem:
Now we just multiply \(\|v\|^{2}\) to both sides and take square roots.
If \(w = 0\) (i.e. \(v\) and \(w\) have no othogonal component, and therefore they are scalar multiples), then this would turn into an equality as desired.
triangle inequality (vectors) See also triangle inequality (complexes)
&ldquo;The length of \(u+v\) is always less than the length of each \(u\) plus \(v\); the third side length is always shorter than the sum of both other sides&rsquo; lengths.&rdquo;
\begin{equation} \|u\| + \|v\| \geq \|u+v\| \end{equation}
\begin{equation} \norm{a} - \norm{b} \leq \norm{a-b} \end{equation}
Notably, the two lines between \(2|\langle u,v \rangle|\) and \(2 \|u\| \|v\|\) holds because of the Cauchy-Schwartz Inequality.
This inequality becomes an equality if \(u\) and \(v\) are a non-negative multiple of the other.
parallelogram equality The sums of squared side lengths of a parallelogram is equal to the sum of the squares of the length of diagonals:
\begin{equation} \|u + v\|^{2} + \|u-v\|^{2} = 2(\|u\|^{2} + \|v\|^{2}) \end{equation}
