cornucopia of analysis — Jemoka Knowledge Base

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’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 &= \langle u,v \rangle - \langle cv, v \rangle \\ &= \langle u,v \rangle - c \langle v,v \rangle \\ &= \langle u,v \rangle - \frac{\langle u,v \rangle}{\|v\|^{2}} \langle v,v \rangle \\ &= \langle u,v \rangle - \frac{\langle u,v \rangle}{\|v\|^{2}} \|v\|^{2} \\ &= 0 \end{align}

Cauchy-Schwartz Inequality \begin{equation} \left(\sum_{k=1}^{n} a_{k} b_{k}\right)^{2} \leq \left(\sum_{k=1}^{n} a_{k}^{2}\right)\left(\sum_{k=1}^{n} b_{k}^{2}\right) \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) “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’ lengths.”

\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}