While vector spaces have additional structure compared to a metric space, a
general vector space has no notion of “length” or “distance.”
Let VV be a vector space over KK. A norm is a function ·:V→R·:V→R such that
- N1.
x
≥
0
∀
x
∈
V
x
≥
0
∀
x
∈
V
- N2. x=0x=0 iff x=0x=0
- N3. αx=|α|x∀x∈Vαx=|α|x∀x∈V, α∈Kα∈K
- N4.
x
+
y
≤
x
+
y
∀
x
,
y
∈
V
x
+
y
≤
x
+
y
∀
x
,
y
∈
V
A vector space together with a norm is called a normed vector space (or
normed linear space).
- V=RNV=RN: x2=∑i=1N|xi|2x2=∑i=1N|xi|2
- V=RNV=RN: x1=∑i=1N|xi|x1=∑i=1N|xi| (“Taxicab”/“Manhattan” norm)
- V=RNV=RN: x∞=maxi=1,...,N|xi|x∞=maxi=1,...,N|xi|
- V=Lp[a,b]V=Lp[a,b], p∈[1,∞)p∈[1,∞): x(t)p=∫ab|x(t)|pdt1/px(t)p=∫ab|x(t)|pdt1/p (The notation Lp[a,b]Lp[a,b] denotes the set of all functions defined on the interval [a,b][a,b] such that this norm exists, i.e., ∥x(t)∥p<∞∥x(t)∥p<∞.)
Note that any normed vector space is a metric space with induced metric d(x,y)=x-yd(x,y)=x-y. (This follows since x-y=x-z+z-y∈x-z+y-zx-y=x-z+z-y∈x-z+y-z.) While a normed vector space “feels like” a metric space, it is important to remember that it actually satisfies a great deal of additional structure.
Technical Note: In a normed vector space we must have (from N2) that x=yx=y if x-y=0x-y=0. This can lead to a curious phenomenon when dealing with continuous-time functions. For example, in L2([a,b])L2([a,b]), we can consider a pair of functions like x(t)x(t) and y(t)y(t) illustrated below. These functions differ only at a single point, and thus ∥x(t)-y(t)∥2=0∥x(t)-y(t)∥2=0 (since a single point cannot contribute anything to the value of the integral.) Thus, in order for our norm to be consistent with the axioms of a norm, we must say that x=yx=y whenever x(t)x(t) and y(t)y(t) differ only on a set of measure zero. To reiterate x=y⇎x(t)=y(t)∀t∈[a,b]x=y⇎x(t)=y(t)∀t∈[a,b], i.e., when we treat functions as vectors, we will not interpret x=yx=y as pointwise equality, but rather as equality almost everywhere.
![A smooth function defined on the interval [-1,1].](03_4.png)
