Prove that the vector space 1 is complete in it’s norm.
Answers
Let X and Y be normed linear spaces, and let B(X,Y) denote the collection of all bounded linear operators from X to Y endowed with the operator norm. Show that B(X,Y) is a normed linear space, and B(X,Y) is a Banach space whenever Y is a Banach space. The vector operations in B(X,Y) are defined pointwise, i.e. (A+B)(x)=Ax+Bx, and (αA)(x)=α(Ax). (Notice that in your case X′=B(X,C) and C is a Banach space)
It is clear that linear operators form a linear space. To show that B(X,Y) is a linear subspace, it is enough to show the closure to addition and scalar multiplication. But these follow easily from the properties of a norm (the fact that the operator norm satisfies all the properties of a norm for bounded functionals is an easy exercise that follows from properties of supremums in [0,∞)) , namely for any A,B∈B(X,Y) and λ∈C
∥A+B∥≤∥A∥+∥B∥<∞
∥λA∥=|λ|⋅∥A∥<∞
Thus, B(X,Y) is a normed linear space.
Now assume that Y is a Banach space. Let {Ai} be a Cauchy sequence in B(X,Y), i.e. ∀ϵ>0, ∃N∈N such that ∀m,n>N, ∥An−Am∥<ϵ. Let x∈X be arbitrary. Let ϵ>0 be arbitrary. If x=0, then
∥Anx−Amx∥=0<ϵ.
If x≠0, choose N such that ∥An−Am∥<ϵ∥x∥. Then by a property of the operator norm, ∀m,n>N,
∥Anx−Amx∥=∥(An−Am)x∥≤∥(An−Am)∥⋅∥x∥<ϵ∥x∥⋅∥x∥=ϵ
Thus, in both cases {Anx} is a Cauchy sequence in Y. Since Y is a Banach space, it is convergent to some element in Y. Call that element Ax, i.e.
limn→∞Anx=Ax
Since x was arbitrary, Ax is defined for any x∈X. Thus, A is a map from X to Y defined by x→Ax. We need to show that A is linear, bounded, and An−→−−n→∞A in the operator norm. Notice that A is linear, since by linearity of An we get that for any x1,x2∈X, λ∈C,
A(x1+x2)=limn→∞An(x1+x2)=limn→∞(Anx1+Anx2)=limn→∞Anx1+limn→∞Anx2=Ax1+Ax2
A(λx1)=limn→∞An(λx1)=limn→∞λ⋅Anx1=λlimn→∞Anx1=λ⋅Ax1
Now recall that Cauchy sequences are bounded. Thus, ∀n, ∥An∥<C for some C∈R. Using this fact, we can see that A is bounded, since by continuity of a norm:
∥A∥=sup∥x∥≤1∥Ax∥=sup∥x∥≤1∥limn→∞Anx∥=sup∥x∥≤1limn→∞∥Anx∥=sup∥x∥≤1lim supn→∞∥Anx∥≤sup∥x∥≤1lim supn→∞(∥An∥⋅∥x∥)≤sup∥x∥≤1C⋅∥x∥=Csup∥x∥≤1∥x∥≤C
Finally, we want to show that An−→−−n→∞A in the operator norm. Let ϵ>0 be arbitrary. Recall that for an arbitrary x∈X, we have
∥Anx−Amx∥≤∥(An−Am)∥⋅∥x∥
Since {An} is Cauchy, choose N big enough such that for all n,m≥N, ∥(An−Am)∥<ϵ. Then the above inequality turns into
∥Anx−Amx∥≤ϵ⋅∥x∥
Now by continuity of a norm, we can take limit on both sides as m goes to infinity to obtain
∥Anx−Ax∥≤ϵ⋅∥x∥
Now taking supremum on both sides over all x such that ∥x∥≤1 yields
sup∥x∥≤1∥Anx−Ax∥≤ϵ
But this is equivalent to saying that for all n≥N,
∥An−A∥≤ϵ
And since ϵ was arbitrary, this implies that
An−→−−n→∞A
in the operator norm. Thus, we conclude that B(X,Y) is a Banach space.
Answer:
Step-by-step explanation:
(i) f(x) ≥ 0 for all x ∈ V
(ii) f(x + y) ≤ f(x) + f(y) for all x, y ∈ V
(iii) f(λx) = |λ|f(x) for all λ ∈ C and x ∈ V
(iv) f(x) = 0 if and only if x = 0
Property (ii) is called the triangle inequality, and property (iii) is called positive homgeneity.
We usually write a norm by kxk, often with a subscript to indicate which norm we are
refering to. For vectors x ∈ R
n or x ∈ C
n
the most important norms are as follows.
• The 2-norm is the usual Euclidean length, or RMS value.
kxk2 =
Xn
i=1
|xi
|
2
1
2
• The 1-norm
kxk1 =
Xn
i=1
|xi
|
• For any integer p ≥ 1 we have the p-norm
kxkp =
Xn
i=1
|xi
|
p
1
p
• The ∞-norm, also called the sup-norm. It gives the peak value.
kxk∞ = max
i
|xi
|
This notation is used because kxk∞ = limp→∞kxkp.
One can show that these functions each satisfy the properties of a norm. The norms are also
nested, so that
kxk∞ ≤ kxk2 ≤ kxk1
This is easy to see; just sketch the unit ball
{ x ∈ R
2
| kxk ≤ 1 }
for each of the norms, and notice that they are nested.
These norms also satisfy pairwise inequalities; for example
kxk1 ≤ nkxk∞ for all x ∈ C