Corollaries to the Equivalence of Norms in a Finite-Dim. Linear Space

+Corollaries to the Equivalence of Norms in a Finite-Dimensional Linear Space]

Recall from the Equivalence of Norms in a Finite-Dimensional Linear Space page that if $X$ is a finite-dimensional linear space then any two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on $X$ are equivalent, that is, there exists positive numbers $C, D > 0$ such that:

(1)
\begin{align} \quad C \| x \|_1 \leq \| x \|_2 \leq D \| x \|_2 \end{align}

For all $x \in X$. We now state some important corollaries to this result.

 Corollary 1: Let $X$ be a normed linear space and let $M$ be a linear subspace of $X$. If $M$ is finite-dimensional then $M$ is closed.
• Proof: Let $M$ be a finite-dimensional subspace of $X$, say $\dim (M) = n$. Let $\{ e_1, e_2, ..., e_n \}$ be a basis of $X$. Then every $m \in M$ can be uniquely written in the form $m = a_1e_1 + a_2e_2 + ... + a_ne_n$, which we will denote by $m = (a_1, a_2, ..., a_n)$. Define a function $\| \cdot \|_{\infty} : M \to [0, \infty)$ for all $m \in M$ by:
(2)
\begin{align} \quad \| m \|_{\infty} = \max \{ |a_k| : 1 \leq k \leq n \} \end{align}
• Then $\| \cdot \|_{\infty}$ is clearly a norm on $M$ and $M$ is complete (if $(m_n)$ is a Cauchy sequence in $M$ then $(m_n)$ will be Cauchy in "coordinate" of $m$ and each coordinate will converge in $\mathbb{R}$ and so $(m_n)$ will converge in $M$). So every Cauchy sequence in $M$ converges in $M$. In particular, every convergent sequence in $M$ is Cauchy and so every convergent sequence converges in $M$ and so $M$ is closed. $\blacksquare$
 Corollary 2:* Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear space. If $X$ is finite-dimensional and $T : X \to Y$ is a linear map then $T$ is bounded. That is, every linear operator whose domain is a finite-dimensional normed linear space is a bounded linear operator.
• Proof: Let $X$ be finite-dimensional with $\dim (X) = n$ and let $\{e_1, e_2, ..., e_n \}$ be a basis of $X$. For each $x \in X$ write $x = a_1e_1 + a_2e_2 + ... + a_ne_n$. Let $\| \cdot \|_{\infty} : X \to \mathbb{R}$ be defined for all $x \in X$ by $\| x \|_{\infty} = \max \{ |a_k| : 1 \leq k \leq n \}$. Since $X$ is finite-dimensional, $\| \cdot \|_X$ and $\| \cdot \|_{\infty}$ are equivalent and so there exists constants $C, D > 0$ such that:
(3)
\begin{align} \quad C \| x \|_X \leq \| x \|_{\infty} \leq D \| x \|_X \end{align}
• For all $x \in X$. Let $M = \sum_{k=1}^{n} \| T(e_k) \|_Y$. Since $T : X \to Y$ is linear we have that for all $x \in X$:
(4)
\begin{align} \quad \| T(x) \|_Y &= \| T(a_1e_1 + a_2e_2 + ... + a_ne_n) \|_Y \\ &= \| a_1T(e_1) + a_2T(e_2) + ... + a_nT(e_n) \|_Y \\ & \leq |a_1| \| T(e_1) \|_Y + |a_2| \| T(e_2) \|_Y + ... + |a_n| \| T(e_n) \|_Y \\ & \leq \sum_{k=1}^n |a_k| \| T(e_k) \|_Y \\ & \leq \sum_{k=1}^{n} \max \{ |a_j| : 1 \leq j \leq n \} \| T(e_k) \|_Y \\ & \leq \sum_{k=1}^{n} \| x \|_{\infty} \| T(e_k) \|_Y \\ & \leq \left ( \sum_{k=1}^{n} \| T(e_k) \|_Y \right ) \| x \|_{\infty} \\ & \leq M \| x \|_{\infty} \\ & \leq MD \| x \|_X \end{align}
• Therefore $T$ is a bounded linear operator. $\blacksquare$