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Suppose $X$ is an Inner Product Space with Inner Product $\langle \cdot, \cdot \rangle: X \to \mathbb{C}$. Then $||\cdot||: X \to \mathbb{R}_{\geq 0}$ defined as $$||\mathbf{x}|| = \sqrt{\langle \mathbf{x}, \mathbf{x} \rangle}$$ is a Norm. This makes $X$ a Normed Vector Space as well.

We must simply show that $|| \cdot ||$ as defined has Codomain $\mathbb{R}_{\geq 0}$ and it satisfies the Norm axioms:

1. By Positive Definiteness, $||\cdot||$ does indeed have Codomain $\mathbb{R}_{\geq 0}$
2. By Positive Definiteness, $||\mathbf{x}|| = \langle \mathbf{x}, \mathbf{x} \rangle = \mathbf{0}$ If and Only If $\mathbf{x} = \mathbf{0}$.
3. Suppose $c \in \mathbb{C}$ and $\mathbf{x} \in X$. Then $$||c \mathbf{x}|| ^ {2} = \langle c \mathbf{x}, c \mathbf{x} \rangle = c \overline{c} \langle \mathbf{x}, \mathbf{x} \rangle = |c|^{2} \langle \mathbf{x}, \mathbf{x} \rangle$$ so $||c \mathbf{x}|| = |c| ||\mathbf{x}||$.
4. Suppose $\mathbf{x}, \mathbf{y} \in X$. Then $$\begin{align*} ||\mathbf{x} + \mathbf{y}|| &= \sqrt{\langle \mathbf{x} + \mathbf{y}, \mathbf{x} + \mathbf{y} \rangle}\\ &= \sqrt{\langle \mathbf{x}, \mathbf{x} \rangle + \langle \mathbf{x}, \mathbf{y} \rangle + \langle \mathbf{y}, \mathbf{x} \rangle + \langle \mathbf{y}, \mathbf{y} \rangle}\\ &=\sqrt{||\mathbf{x}||^{2} + 2 \text{Re}\langle \mathbf{x}, \mathbf{y} \rangle + ||\mathbf{y}||^{2}}\\ &\leq\sqrt{||\mathbf{x}||^{2} + 2 |\langle \mathbf{x}, \mathbf{y} \rangle| + ||\mathbf{y}||^{2}}\\ &\leq\sqrt{||\mathbf{x}||^{2} + 2 ||\mathbf{x}|| ||\mathbf{y}|| + ||\mathbf{y}||^{2}} &\text{(Cauchy Schwarz)}\\ &= ||\mathbf{x}|| + ||\mathbf{y}|| \end{align*}$$

Thus $||\cdot||$ is indeed a Norm, making $X$ a Normed Vector Space. $\blacksquare$