Describe the sets
\begin{equation*}
B_r(0) = \{x \in \R^2: d_{tc}(x,0) \lt r\}.
\end{equation*}
Section 3.1 Norms
When we think about basic notions of open sets in Euclidean space \(\R^n\text{,}\) we typically consider open balls - that is, sets of the form
\begin{equation*}
B_r(a) = \{x \in \R^n: \norm{x - a} \lt r\}
\end{equation*}
using the standard norm Euclidean norm (1.2.2) to define the notion of distance.
Of course, there are other ways to measure the distance between vectors that are consistent with the properties that we associate with metrics. The metric used will change the elements that fall into a “ball of radius \(r\) centered at \(a\)”. For example, on \(\R^2\text{,}\) we can define the taxicab metric
\begin{equation*}
d_{tc}(x, y) = \abs{x_1 - y_1} + \abs{x_2 - y_2}.
\end{equation*}
Checkpoint 3.1.1.
We formalize the notion of metric in the following definition.
Definition 3.1.2.
Let \(X\) be a set. A metric is a map \(d:X\times X \to \R\) which satisfies for all \(x, y, z \in X\)
- \(d(x,y) \geq 0\text{;}\)
- \(d(x,y) = 0\) if and only if \(x = y\text{;}\)
- \(d(x,y) = d(y,x)\text{;}\)
- \(d(x,z) \leq d(x,y) + d(y,z)\) for all \(y\text{.}\)
Our focus is going to be on the analysis of inner product spaces of functions. But many spaces of functions can be equipped with metrics that do not arise from inner products. We’ve already seen \(C[a,b]\text{,}\) the space of continuous functions on \([a,b]\text{,}\) which is an inner product space with inner product (2.1.1). Another natural notion of distance in this space is to measure the maximum modulus of the pointwise distance between the functions:
\begin{equation*}
d_{\sup}(f,g) = \sup_{[a,b]} \abs{f(x) - g(x)}.
\end{equation*}
This is the metric that corresponds to uniform convergence of sequences of functions in \(C[a,b]\) and thus is natural and useful. However, the open sets in this metric do not coincide with the open sets in the inner product topology. In fact, there is no inner product on \(C[a,b]\) that corresponds to the maximum modulus metric.
1
en.wikipedia.org/wiki/Uniform_convergenceBecause we are ultimately interested in analysis involving vector spaces of functions, we will need to consider a broader range of metrics than those strictly arising from inner products. We wish to replicate the properties that a norm inherits from inner product as seen in Theorem 2.2.2 and in Theorem 2.2.6. (These are the properties that make it possible do to analysis!)
Definition 3.1.3.
Let \(V\) be a real or complex vector space. A norm on \(V\) is a map \(\norm{\cdot}: V \to \R\) which satisfies
- \(\norm{x} > 0\) if \(x \neq 0\text{;}\)
- \(\norm{c x} = \abs{c}\norm{x}\) for all \(c \in \F\) and \(x \in V\text{;}\)
- \(\norm{x + y} \leq \norm{x} + \norm{y}\) for all \(x, y \in V\text{.}\)
A normed space is a pair \((V, \norm{\cdot})\) where \(V\) is a real or complex vector space and \(\norm{\cdot}\) is a norm.
We should observe that setting \(c = 0\) into property (2) above gives that \(\norm{0} = 0\) in a normed space.
Checkpoint 3.1.4.
Show that the supremum norm \(\norm{\cdot}_\infty\) on the complex vector space \(C[a,b]\) of all continuous complex valued functions on \([a,b]\) defined by
\begin{equation*}
\norm{f}_\infty = \sup_{[a,b]} \abs{f(x)}
\end{equation*}
satisfies conditions 1-3 in Definition 3.1.3.
A topology on a set \(X\) is a collection of sets \(\tau\text{,}\) called open sets that satisfy the following axioms:
- The empty set and \(X\) are in \(\tau\text{.}\)
- An arbitrary union of members of \(\tau\) belongs to \(\tau\text{.}\)
- A finite intersetion of members of \(\tau\) belongs to \(\tau\text{.}\)
The topology that we’re most used to in analysis is the topology that is generated by open Euclidean balls (such as open intervals in \(\R^n\) for example). Topologies of sets, particularly those arising from metrics, are important in analysis, because they determine which functions are continuous and which sequences converge. For example, recall that a function \(f\) on \(\R\) is continuous if and only if for every open set \(A \subseteq f(X)\text{,}\) we have \(f\inv(A)\) is open. Likewise, we define a convergent sequence by saying that for any epsilon ball about the limit \(x\text{,}\) we have that the tail \((x_n)\) where \(n \geq N\) is contained in the ball.
In the same way that inner products give rise to norms, norms give rise to metrics.
Theorem 3.1.5.
In a normed space \((V, \norm{\cdot})\text{,}\) the (distance) function \(d: V \times V \to \R\) induced by the norm \(\norm{\cdot}\) and defined by
\begin{equation*}
d(x, y) = \norm{x - y}
\end{equation*}
is a translation-invariant metric.
To say that a metric is translation-invariant is to say that \(d(x + z, y+ z) = d(x,y)\) for any \(z \in V\text{.}\)
Proof.
(1): From Definition 3.1.3 (1), we have
\begin{equation*}
d(x, y) = \norm{x - y} > 0.
\end{equation*}
when \(x \neq 0\text{.}\)
(2):
\begin{equation*}
d(x,x) = \norm{x - x} = \norm{0} = 0.
\end{equation*}
(3):
\begin{equation*}
d(x, y) = \norm{x - y} = \abs{-1}\norm{y - x} = \norm{y - x} = d(y,x).
\end{equation*}
(4): From Definition 3.1.3 (3), we have for all \(y \in V\text{,}\)
\begin{align*}
d(x,z) \amp= \norm{x - z}\\
\amp = \norm{x - y + y - z}\\
\amp \leq \norm{x - y} + \norm{y - z}\\
\amp = d(x,y) + d(y,z)
\end{align*}
That \(d\) is translation invariant is straightforward and left as an exercise.
In finite dimensions, this isn’t all that interesting, as we also have that the open sets determined by a norm on a normed space are all the same, regardless of the norms. Norms \(\norm{\cdot}_1, \norm{\cdot}_2\) on a normed space \(V\) are called equivalent if there exist constants \(A, B\) such that for all \(v \in V\text{,}\)
\begin{equation*}
A \norm{v}_1 \leq \norm{v}_2 \leq B \norm{v}_1;
\end{equation*}
that is, every open neighborhood with respect to the metric induced by \(\norm{\cdot}_1\) both contains and is contained by an open neighborhood in \(\norm{\cdot}_2\text{,}\) and so all convergent sequences in one of the norms is also convergent in the other. For a finite dimensional vector space, all norms are equivalent. In infinite dimensions, things are more interesting.
Checkpoint 3.1.6.
Consider the vector space \(C[a,b]\text{.}\) Prove that the sup-norm of example Checkpoint 3.1.4 and the norm induced by the inner product \(\ip{f}{g} = \int f \cc g\) are not equivalent. Hint: it is enough to exhibit a sequence that converges in one of the induced metrics but not the other.
What we’ve developed so far gives a hierarchy of structured spaces, each category enveloping the previous; inner product spaces are normed spaces, and normed spaces are metric spaces. We can go one step further and consider more generally linear spaces that have a toplogy but not necessarily a metric. These spaces are called topological vector spaces, which are foundationally important in their own right, but well beyond the scope of this text. If you continue on your journey into Hilbert space theory, you’ll meet them in functional analysis.
2
en.wikipedia.org/wiki/Topological_vector_spaceAnalysis, particularly at the undergraduate level, can be lampooned as the art of reducing a problem to the triangle inequality and then taking limits. Even this gentle mockery includes a major mathematical idea - namely, that absolute values (and more generally norms) are well-behaved under limits. That is, in normed spaces the basic operators are continuous in the sense that small changes in input objects result in small changes in norm.
Theorem 3.1.7.
In a normed space \((V, \norm{\cdot})\text{,}\) the map \(\norm{\cdot}:V \to \R\) is continuous.
Proof.
Exercise.
Theorem 3.1.8.
In a normed space \((V,\norm{\cdot})\text{,}\) the algebraic operations on \(V\) are continuous. That is, the maps
\begin{gather*}
+: V \times V \to V,\\
\cdot: \F \times V \to V,
\end{gather*}
where \(\F\) denotes the scalar field, are continuous in the respective product topologies.
Proof.
We will show that scalar multiplication is continuous at an arbitrary point \((c, x) \in \F \times V\text{.}\) For any \(k \in \F, y \in V\text{,}\) we have
\begin{align}
\norm{cx - ky} \amp = \norm{cx - kx + kx - ky}\notag\\
\amp \leq \norm{cx - kx} + \norm{kx - ky}\notag\\
\amp = \abs{c-k}\norm{x} + \abs{k}\norm{x - y}.\tag{3.1.1}
\end{align}
Choose \(\eps > 0.\) Recall that we want to show that all nearby points \(ky\) have norm within \(\eps\) of \(c x\text{.}\) We’ll choose neighborhoods of \(c\) and \(x\) in the following way:
\begin{align*}
A_c \amp = \left\{ k \in \F: \abs{k - c} \lt \min\left(\frac{\eps}{2(1 + \norm{x})}, 1\right)\right\},\\
B_x \amp = \left\{y \in V: \norm{y - x} \lt \frac{\eps}{2(1 + \abs{c})}\right\}
\end{align*}
Then \(A_c \times B_x\) is an open neighborhood of \((c, x) \in \F \times V\text{.}\) For \((k,y) \in A_c \times B_x\text{,}\) we know by construction that
\begin{equation*}
\abs{k} \lt 1 + \abs{c}.
\end{equation*}
Then by (3.1.1), we have
\begin{align*}
\norm{cx - ky} \amp \leq \abs{c-k}\norm{x} + \abs{k}\norm{x - y}\\
\amp \leq \frac{\eps}{2(1 + \norm{x})} \norm{x} + (1+ \abs{c})\frac{\eps}{2(1 + \abs{c})}\\
\amp \lt \eps.
\end{align*}
The remaining part of the proof is similar and left as an exercise.
Finally, we note that (unsurprisingly), inner products are continuous on inner product spaces.
Theorem 3.1.9.
In an inner product space \((V, \ip{\cdot}{\cdot})\text{,}\) the inner product is continuous. That is, the map
\begin{equation*}
\ip{\cdot}{\cdot}: V\times V \to \C
\end{equation*}
is continuous in the product topology.
The proof is similar to the proof of Theorem 3.1.8 and is left as an exercise.
