Every Linear Operator on a Fin.-Dim. Normed Linear Space is Bounded

Every Linear Operator on a Finite-Dimensional Normed Linear Space is Bounded

Theorem 1: Let $X$ and $Y$ be normed linear spaces and let $X$ be finite-dimensional. Then every linear operator $T : X \to Y$ is bounded.
  • Proof: Since $X$ is finite-dimensional, say $\dim (X) = n$, there exists a basis $\{ e_1, e_2, ..., e_n \}$ of $X$ such that every element $x \in X$ can be written uniquely in the form:
(1)
\begin{align} \quad x = a_1e_1 + a_2e_2 + ... + a_ne_n \end{align}
  • where $a_1, a_2, ..., a_n \in \mathbb{C}$. Let:
(2)
\begin{align} \quad M = \left ( \sum_{k=1}^{n} \| T(e_k) \|^2 \right )^{1/2} \end{align}
  • Then for each $x \in X$ we have that:
(3)
\begin{align} \quad \| T(x) \| &= \| T (a_1e_1 + a_2e_2 + ... + a_ne_n) \| \\ &= \| a_1 T(e_1) + a_2 T(e_2) + ... + a_n T(e_n) \| \\ & \leq \sum_{k=1}^{n} |a_k| \| T(e_k) \| \end{align}
  • Then by the Cauchy-Schwarz inequality we have that:
(4)
\begin{align} \quad \| T(x) \| & \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \left ( \sum_{k=1}^{n} \| T(e_k) \|^2 \right )^{1/2} \\ & \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \cdot M \end{align}
(5)
\begin{align} \quad \| T(x) \| & \leq M \| x \|_* \end{align}
  • Since $\| \cdot \|$ and $\| cdot \|_*$ are equivalent norms, there exists $C, D \in \mathbb{R}$, $C, D > 0$ such that for every $x \in X$ we have that:
(6)
\begin{align} \quad C \| x \| \leq \| x \|_* \leq D \| x \| \end{align}
  • Therefore, for every $x \in X$ we have that:
(7)
\begin{align} \quad \| T(x) \| & \leq MD \| x \| \end{align}
  • Hence $T$ is bounded. $\blacksquare$
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