Let $X$ be a reflexive [Banach space](/page/Banach%20Space) and let $Y$ be a Banach space. For $T \in \mathcal{L}(X,Y)$, the following are equivalent: \begin{align*} &\text{1. } T \in \mathcal{K}(X,Y),\\ &\text{2. whenever } x_k \rightharpoonup x \text{ in } X, \text{ we have } \|Tx_k-Tx\|_Y \to 0. \end{align*}

Let $V$ be a finite-dimensional inner product space and let $U\in \mathcal{L}(V)$. The following are equivalent: \begin{align*} &U\text{ is unitary},\\ &U^*U=I,\\ &\|Uv\|_V=\|v\|_V\text{ for all }v\in V. \end{align*}

Let $V$ be a finite-dimensional inner product space and let $W\le V$ be a subspace. For every $v\in V$, there exist unique vectors $w\in W$ and $z\in W^\perp$ such that \begin{align*} v=w+z. \end{align*} The vector $w$ is the unique minimizer of $\|v-y\|_V$ over all $y\in W$.

Let $V$ be an $n$-dimensional vector space over $\mathbb{F}$ with basis $(v_1,\ldots,v_n)$. Inner products on $V$ are in bijection with positive-definite Hermitian matrices $G\in \mathbb{F}^{n\times n}$ by \begin{align*} \left(\sum_{j=1}^n a_jv_j,\sum_{i=1}^n b_iv_i\right)_V=b^*Ga. \end{align*}

If $N \subset \mathbb{R}^n$ is a Lebesgue null set and $A \subset N$, then $A$ is Lebesgue measurable and \begin{align*} \mathcal{L}^n(A)=0. \end{align*}

For every Lebesgue measurable set $E \subset \mathbb{R}^n$, [Lebesgue measure](/page/Lebesgue%20Measure) is outer regular on $E$. If $\mathcal{L}^n(E)<\infty$, then Lebesgue measure is inner regular on $E$. Moreover, if $\mathcal{L}^n(E)<\infty$, then for every $\varepsilon>0$ there are an [open set](/page/Open%20Set) $U \subset \mathbb{R}^n$ and a compact set $K \subset \mathbb{R}^n$ such that \begin{align*} K \subset E \subset U, \qquad \mathcal{L}^n(U \setminus K) < \varepsilon. \end{align*}

If $E \subset \mathbb{R}^n$ is Lebesgue measurable and $a \in \mathbb{R}^n$, then $E+a$ is Lebesgue measurable and \begin{align*} \mathcal{L}^n(E+a)=\mathcal{L}^n(E). \end{align*}

The collection of Lebesgue measurable subsets of $\mathbb{R}^n$ is a $\sigma$-algebra containing all open subsets of $\mathbb{R}^n$. The measure $\mathcal{L}^n$ is countably additive on this $\sigma$-algebra. If $(E_k)_{k=1}^\infty$ are pairwise disjoint Lebesgue measurable sets, then \begin{align*} \mathcal{L}^n\left(\bigcup_{k=1}^\infty E_k\right) = \sum_{k=1}^\infty \mathcal{L}^n(E_k). \end{align*} Moreover, $\mathcal{L}^n(\varnothing)=0$ and $\mathcal{L}^n(E) \le \mathcal{L}^n(F)$ whenever $E \subset F$ are Lebesgue measurable.