This is a recap on basic definitions and theorems in functional analysis. Mainly from a youtube course taught by Claudio Landim.

  • If $X$ is a linear space (or vector space), $Y \subseteq X$ is a linear subspace if $\forall x, y \in Y$ and $\forall \alpha \in \mathbb{K}$ where $\mathbb{K}$ is the underlying field, we have $\alpha x + y \in Y$.

  • If \(\{Y_{\theta} ~:~ \theta \in I\}\) is a family of linear subspaces in $X$, then \(\bigcap_{\theta \in I} Y_{\theta}\) is a linear subspace of $X$.

  • A family of sets \(\{Y_{\theta} ~:~ \theta \in I\}\) is totally ordered if $\forall \theta_1, \theta_2 \in I$ either $Y_{\theta_1} \subseteq Y_{\theta_2}$ or $Y_{\theta_2} \subseteq Y_{\theta_1}$.

  • If \(\{Y_{\theta} ~:~ \theta \in I\}\) is a family of linear subspaces in $X$ which is totally ordered, then \(\bigcup_{\theta \in I} Y_{\theta}\) is a linear subspace of $X$.

  • Let the set $S \subseteq X$ be a subset of a linear space, then the linear span of $S$ denoted by $LS(S)$ is the defined as the intersection of all linear subspaces of $X$, i.e. \(\bigcap Y_{\theta}\) where $Y_{\theta}$ is a linear subspace of $X$. It can then be proved that the linear span of $S$ is the smallest linear subspace which contains $S$.

  • It can be proved that the linear span of $S$ is formed by considering all linear combinations of elements of $S$, i.e. \(LS(S) = \left\{ \sum_{j=1}^n \alpha_j x_j, n \ge 1, \alpha_j \in \mathbb{K}, x_j \in S \right\}\). This shows that all the elements of the linear span can be written as the finite linear combination of the elements of $S$.

  • Let $Y$ be a linear subspace of a set $X$ over a field $\mathbb{K}$, then we can prove that the relation $\sim$ defined as $x \sim y$ if $x - y \in Y$ is an equivalence relation. If we define $[x] = \{ y \in X ~:~ y \sim x\}$, then the quotient space of $X$ modulo $Y$ is defined as \(X \lvert Y \triangleq \{[x] ~:~ x \in X\}\). It can be proved that the summation $[x]+[y] = [x + y]$ and the multiplication $\alpha [x] = [\alpha x]$ are well defined and the quotient space is a linear space under this summation and multiplication.

  • Let $X^{\star}, X$ be two spaces over a field $\mathbb{K}$, a map $T : X \to X^{\star}$ is a linear map if $\forall x, y \in X$ we have $T(x+y) = T(x) + T(y)$ and $\forall x \in X, \forall \alpha \in \mathbb{K}$ we have $T(\alpha x) = \alpha T(x)$. Linear spaces $X^{\star}$ and $X$ are isomorph if there exists a linear map $T : X \to X^{\star}$ which is a bijection.

  • If $X$ is a space and $T : X \to X^{\star}$ is a linear map from $X$ and $Y \subseteq X$ is a linear subspace of $X$ and $Y^{\star} \subseteq X^{\star}$ is a linear subspace of $X^{\star}$, then
    • \(T(Y) = \{T(x) ~:~ x \in Y \}\) is a linear subspace.
    • \(T^{-1}(Y^{\star}) = \{ x \in X ~:~ T(x) \in Y^{\star} \}\) is a linear subspace.
  • Any linear subspace is a convex set.

  • If $Y_1$ and $Y_2$ are convex subsets of a set $X$, then \(Y_1 + Y_2 = \{y_1+y_2 ~:~ y_1 \in Y_1, y_2 \in Y_2~ \}\) is a convex set.

  • If \(\{Y_{\theta} ~:~ \theta \in I\}\) is a family of convex sets, then then \(\bigcap_{\theta \in I} Y_{\theta}\) is a convex set.

  • If \(\{Y_{\theta} ~:~ \theta \in I\}\) is a totally ordered family of convex sets, then then \(\bigcup_{\theta \in I} Y_{\theta}\) is a convex set.

  • If $X$ is a space and $T : X \to Y$ is a linear map and $K \subseteq X$ is a convex set, then $T(K)$ is a convex set.

  • If $X$ is a space and $T : X \to Y$ is a linear map and $K \subseteq Y$ is a convex set, then $T^{-1}(K)$ is a convex set.

  • If $S \subset X$, and \(\{K_{\theta} ~:~ \theta \in I\}\) is the family of all convex sets that contain $S$, i.e. $S \subseteq K_{\theta}$, then convex hull of $S$ is the intersetion of all these convex sets, i.e. \(cv(S) \triangleq \bigcap_{\theta \in I} K_{\theta}\). It can be proved that the convex set is the smallest convex set which contains $S$. Further it can be proved that the convex hull of $S$ is equal to the set of all convex combinations of elements of $S$, i.e. \(cv(S) = \left\{ \sum_{j=1}^n \alpha_j x_j ~:~ n \ge 1, x_j \in S, \alpha_j \in [0,1], \sum_{j=1}^n {\alpha_j} =1 \right\}\).