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 $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $X$ be finite-dimensional. Then every linear operator $T$ from $X$ 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:

\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:

\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:

\begin{align} \quad \| T(x) \|_Y &= \| T (a_1e_1 + a_2e_2 + ... + a_ne_n) \|_Y \\ &= \| a_1 T(e_1) + a_2 T(e_2) + ... + a_n T(e_n) \|_Y \\ & \leq \sum_{k=1}^{n} |a_k| \| T(e_k) \|_Y \end{align}

- Then by the Cauchy-Schwarz inequality we have that:

\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) \|_Y^2 \right )^{1/2} \\ & \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \cdot M \end{align}

- From the Equivalence of Norms in a Finite-Dimensional Linear Space page we have that:

\begin{align} \quad \| T(x) \|_Y & \leq M \| x \|_* \end{align}

- Since $\| \cdot \|_X$ 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:

\begin{align} \quad C \| x \|_X \leq \| x \|_* \leq D \| x \|_X \end{align}

- Therefore, for every $x \in X$ we have that:

\begin{align} \quad \| T(x) \| & \leq MD \| x \|_X \end{align}

- Hence $T$ is bounded. $\blacksquare$