Definition 2.2.1.
The norm of a vector \(x\) in an inner product space \(V\text{,}\) denoted \(\norm{x}\) is defined to be
\begin{equation*}
\norm{x} = \ip{x}{x}^{1/2}
\end{equation*}
Section 2.2 Inner products, norms, and metric spaces
A metric is a function that measures distance in a space. The most obvious example of a metric that we encounter in elementary mathematics is the Euclidean metric, which arises from the expression for the magnitude of a vector in, e.g., \(\R^2\text{:}\)
\begin{equation*}
\abs{x} = \sqrt{x_1^2 + x_2^2} = \ip{x}{x}^{1/2}.
\end{equation*}
This relation underlies the distance formula
\begin{equation*}
d(x,y) = \abs{x - y} = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}
\end{equation*}
which is the Euclidean metric on \(\R^2\text{.}\) This relationship between magnitude (in terms of an inner product) and distance between vectors can be generalized to any vector space that has an inner product.
That is, in an inner product space, the inner product gives a natural method for evaluating magnitude of a vector. So with the dot product on \(\C^n\text{,}\) we get
\begin{equation*}
\norm{x} = \sqrt{\abs{x_1}^2 + \ldots + \abs{x_n}^2}.
\end{equation*}
Using the inner product set out in Checkpoint 2.1.2, for \(f \in C[0,1]\text{,}\)
\begin{equation*}
\norm{f} = \left( \int_0^1 \abs{f(t)}^2 \, \dd t \right)^{1/2}.
\end{equation*}
(At this point, you might be suspicious that there is a relationship between functions that are square integrable and sequences that are square summable. Keep that suspicion close!)
Some basic facts about the behavior of norms induced by inner products follow.
Theorem 2.2.2.
Let \(V\) be a inner product space. For any \(x \in V\) and \(c \in \C\text{,}\)
- \(\norm{x} \geq 0\) and \(\norm{x} = 0\) if and only if \(x = 0\text{;}\)
- \(\norm{cx} = \abs{c}\norm{x}\text{.}\)
Checkpoint 2.2.3.
Prove Theorem 2.2.2.
In Euclidean space, we get a suite of geometric relations that we can use to work with vectors. In a general inner product space, we recover some of the same. In \(\R^2\text{,}\) the dot product contains information about the angle \(\theta\) between two vectors:
\begin{equation*}
\ip{x}{y} = \norm{x}\norm{y} \cos \theta.
\end{equation*}
In the complex or abstract setting, the notion of angle between vectors no longer makes sense, but the relationship of the magnitudes still holds in the form of the Cauchy-Schwarz inequality.
Theorem 2.2.4.
For any \(x, y\) in an inner product space \(V\text{,}\)
\begin{equation}
\abs{\ip{x}{y}} \leq \norm{x}\norm{y}.\tag{2.2.1}
\end{equation}
Equality holds if and only if \(x\) and \(y\) are linearly dependent.
Proof.
First, suppose that \(x\) and \(y\) are linearly dependent so that \(x = c y\) for some \(c \in \C\text{.}\) Then by Theorem 2.2.2 and Definition 2.1.1, the left and right sides of (2.2.1) are both equal to \(\abs{c}\norm{y}\text{.}\)
On the other hand, if \(x, y\) are linearly independent, then for any \(c \in \C\text{,}\) it must be that \(x + c y \neq 0\text{.}\) Then
\begin{align*}
0 \amp \lt \ip{x+cy}{x+cy}\\
\amp = \ip{x}{ x + c y} + \ip{c y}{x + cy}\\
\amp = \ip{x}{x} + \cc{c}\ip{x}{y} + c\ip{y}{x} + c \cc{c} \ip{y}{y}\\
\amp = \norm{x}^2 + \cc{c}\ip{x}{y} + c \cc{\ip{x}{y}} + \abs{c}^2 \norm{y}^2\\
\amp = \norm{x}^2 + 2 \RE\left[ \cc{c} \ip{x}{y} \right] + \abs{c}^2 \norm{y}^2.
\end{align*}
To get the real part of the complex number out of the quadratic expression, let \(\omega\) be a unimodular constant so that \(\cc \omega \ip{x}{y} = \abs{\ip{x}{y}}\text{.}\) Now let \(c = t\omega\) and substitute to conclude that for all \(t \in \R\text{,}\)
\begin{equation*}
0 \lt \norm{x}^2 + 2t\abs{\ip{x}{y}} + t^2 \norm{y}^2.
\end{equation*}
That is, a quadratic polynomial in \(t\) is strictly positive. This can only happen if the discriminant is negative, which gives
\begin{equation*}
4 \ip{x}{y}^2 - 4 \norm{x}^2\norm{y}^2 \lt 0,
\end{equation*}
and so
\begin{equation*}
\ip{x}{y} \lt \norm{x}\norm{y}.
\end{equation*}
Checkpoint 2.2.5.
Prove that for any \(f\in C[0,1]\text{,}\)
\begin{equation*}
\abs{\int_0^1 f(t) \sin \pi t\, \dd t} \leq \frac{1}{\sqrt{2}} \left(\int_0^1 \abs{f(t)}^2 \, \dd t \right)^{1/2},
\end{equation*}
and find the functions \(f\) for which equality holds.
The Cauchy-Schwarz inequality is the first of a murderers row of important geometric inequalities. The next is probably the single most important inequality in mathematical analysis.
Theorem 2.2.6. Triangle inequality.
For any \(x, y\) in \(V\text{,}\)
\begin{equation*}
\norm{x + y} \leq \norm{x} + \norm{y}.
\end{equation*}
Proof.
A typical trick is to avoid square roots by using the squares of norms when working with inner products. Hence,
\begin{align*}
\norm{x + y}^2 \amp = \norm{x}^2 + 2 \RE \ip{x}{y} +\norm{y}^2\\
\amp \leq \norm{x}^2 + 2 \abs{\ip{x}{y}} + \norm{y^2}\\
\amp \leq \norm{x}^2 + 2 \norm{x}\norm{y} + \norm{y}^2\\
\amp = (\norm{x} + \norm{y})^2.
\end{align*}
Monotonicity of the square root gives the result.
Checkpoint 2.2.7.
Use Theorem 2.2.6 to show that \(\ell^2\) is closed under addition.
Theorem 2.2.8. Parallelogram identity.
For \(x, y \in V\text{,}\)
\begin{equation}
\norm{x + y}^2 + \norm{x - y}^2 = 2 \norm{x}^2 + 2 \norm{y}^2.\tag{2.2.2}
\end{equation}
Proof.
Using Definition 2.2.1, it is straightforward to compute that
\begin{align*}
\norm{x + y}^2 \amp = \norm{x}^2 + \ip{x}{y} + \ip{y}{x} + \norm{y}^2,\\
\norm{x - y}^2 \amp = \norm{x}^2 - \ip{x}{y} - \ip{y}{x} + \norm{y}^2.
\end{align*}
Adding the expressions above gives the result.
So far, we’ve used the inner product to compute a norm on an inner product space. If on the other hand we know how to compute norms, we can recover the inner product, should it exist (e.g. if we know that we are in an inner product space).
Theorem 2.2.9. Polarization identity.
For \(x, y\) in an inner product space \(V\text{,}\)
\begin{equation}
\ip{x}{y} = \frac{1}{4} \left(\norm{x + y}^2 - \norm{x - y}^2 + i \norm{x - iy}^2 - i\norm{x + iy}^2\right)\tag{2.2.3}
\end{equation}
Checkpoint 2.2.10.
Prove Theorem 2.2.9.
(A larger result, not proven here, is due to Frechet, von Neumann, and Jordan and states that in a normed space \((V, \norm{\cdot})\text{,}\) if the parallelogram law holds, then there is an inner product on \(V\) so that \(\norm{x} = \ip{x}{x}\text{.}\) We will see soon that while every inner product space carries a norm, the converse statement is not true. )
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en.wikipedia.org/wiki/Polarization_identity#Theorem
