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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
1 Normed vector spaces
Mods analysis on the real line derives many of its basic statements (algebra of limits, of continuous functions etc) just using a few basic properties of the modulus of real numbers. Later the very same properties - this time of the length (or modulus of a vector) are used for 2,3 or several variables to obtain similar results. Here we will extract these properties in an axiomatic way, in order to do analysis on any linear space.
1.Definition Let X be a vector space over a field F, where F = R or C. A norm on X is a map x 7 → ‖x‖ of X into [0, ∞) with the following properties:
(i) ‖x‖ ≥ 0 for all x ∈ X (positivity),
(ii) ‖x‖ = 0 ⇐⇒ x = 0 (non degenerate),
(iii) ‖λx‖ = |λ|‖x‖ for all λ ∈ F, x ∈ X ((1)-homogeneity),
(iv) ‖x + y‖ ≤ ‖x‖ + ‖y‖ for all x, y ∈ X (triangle inequality).
Then (X, ‖ · ‖) is a normed vector space.
We often just say that X is a normed vector space if the norm is clear, and on the other hand sometimes write ‖x‖X in order to specify the norm explicitely.
Useful fact (“second triangle inequality”): |‖x‖ − ‖y‖| ≤ ‖x − y‖
Proof We have from the 1.(iv) that ‖x‖ ≤ ‖x − y‖ + ‖y‖ and hence ‖x‖ − ‖y‖ ≤ ‖x − y‖. Interchanging x and y we get ‖y‖ − ‖x‖ ≤ ‖y − x‖ = ‖(−1)(x − y)‖ = | − 1 |‖x − y‖ = ‖x − y‖. Since t ≤ s and −t ≤ s implies |t| ≤ s, the proof is finished.
2.Example In the following we will see, that often the triangle inequality is the property which is hardest to show. But attention has to be paid to all four!
- X = Fm. There are many norms and this one space. For x = (x 1 ,... , xm),
‖x‖ 2 =
( (^) m ∑
1
|xj |^2
(Euclidean norm),
‖x‖ 1 =
∑^ m
1
|xj |, (Manhattan norm)
‖x‖∞ = max j |xj |. (Chess King’s norm)
The names attached to the second and third norm naturally appear if one understands the ‖x‖ as the distance, i.e. the length of the travel) from x the origin and now imagines that the travel happens (for m = 2) in the situation mentioned. (moving along |x 1 | (streets) + |x 2 | (avenues) blocks, resp. every neihgbouring chess field being in distance one for the king). Let us verify the properties 1.(i)–(iv), in order to get some experience in working with norms. Of course, (i) is trivial since sums and maxima of non-negative numbers as well as any square root are non-negative. Next, (ii) follows easily from the useful inequality
0 ≤ |xj | ≤ ‖x‖∞‖x‖ 2 ≤ ‖x‖ 1 for all x ∈ Fm, j ≤ m,
where the 3rd and 4th “≤ ” become obvious when taking the square. In order to verify (iii), let us first notice (for later usage), that it would follow from (i) and the weaker (iii)’ ‖λx‖ ≤ |λ|‖x‖ for all λ ∈ F, x ∈ X Indeed, this gives 0 ≥ ‖ 0 x‖ = 0 = | 0 |‖x‖ as required if λ = 0 and for λ 6 = 0 we use
‖x‖ =
λ
(λx)
λ
∣ ‖λx‖^ =^
|λ|
‖λx‖ ⇒ |λ|‖x‖ ≤ ‖λx‖,
which together with (iii)’ gives all of (iii). Now (iii)’ is easily established since |λxj | = |λxj ||xj | ≤ |λ|‖x‖∞ gives after taking the maximum over j = 1 ,... , m ‖λx‖∞ ≤ |λ|‖x‖∞. For ‖ · ‖ 1 and ‖ · ‖ 2 the homogeneity follow from a straightforward calculation. Also the ∆-inequality is easy in the non- euclidean cases, as
‖x + y‖ 1 =
∑^ m
j=
|(x + y)j | =
∑^ m
j=
|xj + yj | ≤
∑^ m
j=
|xj | + |yj | = ‖x‖ 1 + ‖y‖ 1 ,
More generally, for any real 1 ≤ p < ∞ (thus p not necessarily an inte- ger!), there is a norm given by:
‖x‖p =
( (^) m ∑
1
|xj |p
) 1 /p .
If p < 1 and the dimension m > 1, the triangle inequality fails.
If p ≥ 1 the triangle inequality is known as Minkowski’s inequality. Its proof is based on Hoelder’s inequality (a generalization of the Cauchy- Schwarz inequality |〈x, y〉| ≤ ‖x‖ 2 ‖y‖ 2 ) ∣ ∣ ∣ ∣ ∣
∑^ m
j=
xj y¯j
≤ ‖x‖p‖y‖q where
p
q
which is in turn a consequence of Young’s inequality
sαt^1 −α^ ≤ αs + (1 − α)t, if s, t ≥ 0 and α ∈ (0, 1).
See Kreyszig, Section 1.2 or the problem sheet for details.
- Let ℓ^1 be the space of all real (or complex) sequences x = (xj )j≥ 1 such that
j |xj^ |^ <^ ∞. Put
‖x‖ 1 =
∑^ ∞
1
|xj |.
More generally, for 1 ≤ p < ∞, let ℓp^ be the space of all sequences such that
j |xj^ | p (^) < ∞, with ‖x‖p = (∑∞ 1 |xj^ | p)^1 /p. Let ℓ∞^ be the space of all bounded sequences with ‖x‖∞ = supj |xj |. These all are subspaces of the space F∞, the infinite dimensional space of all scalar sequences which intuitively is too big to do analysis on it (to find a good norm on it). It should be noticed that
ℓ^1 ⊂ ℓp^ ⊂ ℓq^ ⊂ ℓ∞^ if 1 < p ≤ q < ∞,
so different norms have different ”natural” domains of definition - an infinite dimensional phenomenon whereas on Fm^ we had many different norms. When it comes to verifying positivity, non-degeneracy and 1-homogeneity we can use the arguments explained in great detail in 1.. Also the ∆- inequality is derived from the finite dimensional case by the following ap- proximation argument. For
m ∈ { 1 , 2 ,.. .}, x = (xj ) ∈ ℓp^ let Cm((xj )) = (x 1 , x 2 ,... , xm) ∈ Fm.
Then
‖(xj ) + (yj )‖p = lim m→∞ ‖Cm((xj ) + (yj )‖p
= lim m→∞
‖Cm((xj )) + Cm((yj ))‖p
≤ lim m→∞ (‖Cm((xj ))‖p + ‖Cm((yj ))‖p) , by Minkoski’s inequality
= ‖(xj )‖p + ‖(yj )‖p.
- We already used the ‖ · ‖∞-norm
Fm^ ∼= F{^1 ,...,m}^ ∼= {f : { 1 ,... , m} → F}
and on the space ℓ∞^ of bounded sequences in FN^ ∼= {f : N → F}. It is just natural to go one step further and consider for a general non-empty set Ω the space Fb(Ω) of all bounded F-valued functions on Ω. This means
Fb(Ω) = {f : Ω → F : ∃R∀t ∈ Ω : |f (t)| ≤ R},
which is again equipped with the supremums norm ‖f ‖∞ = supt∈Ω |f (t)|. Recall that by definition of the supremum ‖f ‖∞ is the minimum of all the R three lines above. The norm properties of ‖ · ‖∞ can be checked as in
- Recall/double check that ‖fn − f ‖∞ → 0 iff the functions fn converge uniformly to f. An important space of bounded functions occurs if Ω is a compact topo- logical space, and we consider C(Ω) the space of all continuous functions f : Ω → F. Since continuous scalar functions on a compact space are always bounded, we have then C(Ω) ⊂ Fb(Ω) and therefore
‖f ‖∞ = sup t∈Ω
|f (t)|
defines again the sup-norm — also called uniform norm), since uniform con- vergence is the right kind of convergence to preserve continuity.
If Ω = [a, b], we could obtain another norm on C[a, b] by
‖f ‖ 1 =
∫ (^) b
a
|f (t)| dt.
Properties (i),(iii) and (iv) are clear. It needs, however, some care to check that ‖ · ‖ 1 is non-degenerate. But indeed if f ∈ C([0, 1]) is not the zero function, then there is some t ∈ [0, 1] such that f (t) 6 = 0, so we find an ε > 0
and thus f ∈ N and [f ] = 0L 1 gives (ii) and therefore defines a norm on L^1 (R). We are often sloppy and fail to distinguish notationally between L 1 (R) and L^1 (R), or between f and [f ]. This helps our intuition, but we occasionally have to be careful to include “a.e.” in our statements. For instance the question whether [f ] ⊂ C[0, 1] makes no sense since for every f there is a discontinuous g with g = f a.e. It makes sense to ask if [f ] ∩ C[0, 1] 6 = ∅, if ”[f ] contains a continuous representative”.
Similarly, we can define L^1 (0, 1) L^1 (R^2 )
etc. Also, for 1 ≤ p < ∞, we can define
Lp(R)
in a similar way, as (equivalence classes of) measurable functions f such that |f |p^ is integrable, with
‖f ‖p =
|f |p
) 1 /p .
The fact that this expression satisfies the ∆-inequality is again called Minkowski inequality and concluded from an integral version of Hoelder’s inequality de- rived again from Young’s inequality (Compare with Q4 on Sheet 1). The underlying common idea of the construction (to factorize out the subspace N of functions that play no role for the integration, i.e. are neglegible for candidate for a norm we already have) is implemented in full generality in Q2 of Sheet 1.
Finally, we consider the Lebesgue space of ”essentially bounded” func- tions L∞(R). This is the space bounded measurable functions f : R → F which are a.e. equal to a bounded function g : R → F, with the norm
‖f ‖∞ = inf{α > 0 : |f (t)| ≤ α a.e.}.
We notice that the definition is close in spirit to the one for Fb^ involving the R’s. The situation is different from the one for Lp, p < ∞ since functions in N = {f : f = 0a.e.} are not at all negligible from the viewpoint of the natural candidate ”supremums norm”, and the general way how to construct a norm on the quotient space is introduced only in the 4th year. We can, however, easily
analyse the situation at hand. So let f = g a.e. be measurable and g ∈ Fb. We define the measurable (why !?) sets At = {x : |f (x)| > t}, for t ≥ 0 and B(f ) = {t : At is a null set}. Since At ⊂ {x : f (x) 6 = g(x)} ∪ {x : |g(x)| > t}, we have ‖g‖b F ∈ B(f ). Clearly, At ⊂ As if s < t and At = ∪∞ k=1At+ 1 k
.
From this we see that B(f ) is an interval reaching to +∞ and containing ‖g‖Fb. Moreover, ‖f ‖∞ = ‖f ‖L∞^ = inf B(f ) belongs itself into B(f ) since for each k clearly ‖f ‖L∞ + (^1) k ∈ B(f ) and thus A‖f ‖L∞ is a null set again. So we could have used min instead of inf in the definition above. Summarizing, we have see ‖f ‖L∞ ≤ ‖g‖Fb for each g ∈ Fb^ that is equal to f a.e. We now claim that equality is attained. Indeed, define
g(x) =
{ 0 x ∈ A‖f ‖L∞ f (x) else.
Since ‖f ‖L∞^ ∈ B(f ) we see that g = f a.e. and clearly ‖g‖Fb = supx |g(x)| ≤ ‖f ‖L∞ because x /∈ A‖f ‖L∞ implies |g(x)| = |f (x)| ≤ ‖f ‖L∞. This shows
‖[f ]‖L∞^ = min g∈[f ]
‖g‖Fb ,
and this will also be the general construction of a quotient norm used in the 4th year (but the min to be replaced by a inf).
Metric, Topology
Any norm on X induces a metric on X:
d(x, y) = ‖x − y‖,
i.e. the distance between two points (e.g. vectors) is the “length” of their difference. We can therefore talk about topological properties in X. Thus:
- converges xn → x ⇐⇒ ‖xn − x‖ → 0,
- A subset Y of X is closed if and only if: Whenever xn ∈ Y , x ∈ X, ‖xn−x‖ → 0, then x ∈ Y ,
- T : X → X is continuous if and only if: Whenever xn → x, then T xn → T x.
- A subset Y of X is open if and only if: For all y ∈ Y there is ε > 0 s.t. ‖x − y‖ < ε ⇒ x ∈ Y.
Finally, to obtain 4. we use induction in N and 1. together with 2. to see that
(λ 1 ,... , λN ), (x 1 ,... , xN ) 7 →
∑^ N
j=
λj xj is continuous on FN^ × XN^ → X.
Now we use 3. to infer that the composition of this map with the norm y → ‖y‖ is still continuous and obtain the conlusion.
