Let $X\in L^1(\Omega,\mathcal F,\mathbb P)$. If $\mathcal G=\{\varnothing,\Omega\}$, then \begin{align*} \mathbb E[X\mid\mathcal G]=\mathbb E[X] \end{align*} $\mathbb P$-a.e. If $X$ is $\mathcal G$-measurable, then \begin{align*} \mathbb E[X\mid\mathcal G]=X \end{align*} $\mathbb P$-a.e.

Let $X$ be an infinite-dimensional [Banach space](/page/Banach%20Space) and let $T \in \mathcal{L}(X)$ be compact. Then $0 \in \sigma(T)$.

Let $X$, $Y$, and $Z$ be normed vector spaces. If $T: X \to Y$ is compact, $A: Z \to X$ is bounded and linear, and $B: Y \to Z$ is bounded and linear, then $T A: Z \to Y$ and $B T: X \to Z$ are compact.

Let $X$ be a [Banach space](/page/Banach%20Space), let $Y$ be a [normed vector space](/page/Normed%20Vector%20Space), and let $T: X \to Y$ be compact. If $x_k \rightharpoonup x$ weakly in $X$, then \begin{align*} \|Tx_k - Tx\|_Y &\to 0. \end{align*}

Let $X$ be an infinite-dimensional complex [Banach space](/page/Banach%20Space) and let $T \in \mathcal{L}(X)$ be compact. If $\lambda \in \sigma(T)$ and $\lambda \ne 0$, then $\lambda$ is an eigenvalue of $T$. Moreover, the eigenspace $\ker(T-\lambda I)$ is finite-dimensional.

Let $(X,\|\cdot\|_X)$ be a [normed vector space](/page/Normed%20Vector%20Space), let $(Y,\|\cdot\|_Y)$ be a [Banach space](/page/Banach%20Space), and let $T:X\to Y$ be a [linear map](/page/Linear%20Map). Then $T$ is compact, meaning that $\overline{T(A)}^{\,Y}$ is compact in $Y$ for every bounded subset $A\subset X$, if and only if for every bounded sequence $(x_k)_{k\in\mathbb N}$ in $X$, the sequence $(Tx_k)_{k\in\mathbb N}$ has a subsequence converging in $Y$.

Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space), let $Y$ be a normed [vector space](/page/Vector%20Space), and let $T: X \to Y$ be a bounded finite-rank linear operator. Then $T$ is compact.

Let $X$ and $Y$ be normed vector spaces over the same field $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$, and let $T: X \to Y$ be a [linear map](/page/Linear%20Map). Define the closed unit ball of $X$ by \begin{align*} B_X = \{x \in X : \|x\|_X \leq 1\}. \end{align*} Then $T$ is a compact linear operator if and only if $T(B_X)$ is relatively compact in $Y$, meaning that $\overline{T(B_X)}$ is compact in $Y$.

Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space) and let $Y$ be a [Banach space](/page/Banach%20Space). Suppose $T_m: X \to Y$ is compact for every $m \in \mathbb{N}$ and \begin{align*} \|T_m - T\|_{\mathcal{L}(X,Y)} &\to 0 \end{align*} for a [bounded linear operator](/page/Bounded%20Linear%20Operator) $T: X \to Y$. Then $T$ is compact.

Let $X$ and $Y$ be normed vector spaces. If $S,T: X \to Y$ are compact and $\alpha,\beta$ are scalars, then $\alpha S + \beta T$ is compact.