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1. Theorem .1 For a linear operator T : X → Y between 2 normed spaces, the following statements are
equivalent:
1. T is a bounded operator
2. There exists M ≥ 0 such that ||T (x)|| < M · ||x|| holds for all x ∈ X.
3. T is continuous at 0.
4. T is continuous.
Proof (1) =⇒ (2). We have seen that ||T (x)|| ≤ ||T || · ||x|| holds for all x ∈ X. This if T , is a bounded
operator, then (2) holds for any choice of the real number M with M ≥ ||T ||. (2) =⇒ (3). Clearly,
lim ||xn|| = 0 combined with the inequality ||T (xn)|| ≤ M · xn =⇒ lim ||T (xn)|| = 0. That is T is continuous
at 0.
(3) =⇒ (4). The uniform continuity of T follows immediately from the idenity that
||T (x)− T (y)|| = ||T (x− y)|| and the simple fact that limxn = x holds in a normed space iff lim(xn − x) = 0.
(4) =⇒ (1). Assume by the way of contradiction that ||T || = ∞. Then there exists a sequence {xn} of X
with ||xn|| = 1 and ||T (xn)|| ≥ n for each n. Let yn = xnn for each n and note that ||yn|| = 1/n implies
lim yn = 0. But then by continuity of T we must have limT (yn) = 0, contrary to ||T (yn) = n−1 · ||T (xn)|| ≥ 1
for each n. Thus ||T || < ∞ holds.
Note. From (2) it follows that T is Lipschitz continuous because
||T (x)|| ≤ M · ||x||
=⇒ ||T (x− y)|| ≤ M · ||x− y||
=⇒ ||T (x)− T (y)|| ≤ M · ||x− y||
1. You can refer to the proof here at this link matrixeditions.com/FA.Chap3.1-4.pdf.
1. Theorem .2 Let E and F two topological vector spaces, and T : E → F a linear map. If E is finite
dimensional, then T is continuous.
Proof First, if (e1, . . . , en) is a basis of E, then any set of n+ 1 vectors of T (E) is linearly dependent, so T (E)
has a dimension ≤ n+ 1. Let k be the dimension of T (E), and (v1, . . . , vk) a basis of this space. We can write
for any x ∈ E: T (x) =
∑k
i=1 ai(x)vi and since vi is a basis each ai is linear. We have to show...