Measure, Integration & Real Analysis, 2021a

Section 6E Consequences of Baire’s Theorem 187
The surjectivity and linearity of T imply that
∞⋃
∞⋃
W = T(kB) = kT(B).
k=1
Thus W = ⋃ ∞
k=1
kT(B). Baire’s Theorem [6.76(a)] now implies that kT(B) has a
nonempty interior for some k ∈ Z + . The linearity of T allows us to conclude that
T(B) has a nonempty interior.
Thus there exists g ∈ B such that Tg ∈ int T(B). Hence
k=1
0 ∈ int T(B − g) ⊂ int T(2B) =int 2T(B).
Thus there exists r > 0 such that B(0, 2r) ⊂ 2T(B) [here B(0, 2r) is the closed ball
in W centered at 0 with radius 2r]. Hence B(0, r) ⊂ T(B). The definition of what it
means to be in the closure of T(B) [see 6.7] now shows that
h ∈ W and ‖h‖ ≤r and ε > 0 =⇒ ∃f ∈ B such that ‖h − Tf‖ < ε.
For arbitrary h ̸= 0 in W, applying the result in the line above to
6.82 h ∈ W and ε > 0 =⇒ ∃f ∈ ‖h‖
r
B such that ‖h − Tf‖ < ε.
r h shows that
‖h‖
Now suppose g ∈ W and ‖g‖ < 1. Applying 6.82 with h = g and ε = 1 2
, we see
that
there exists f 1 ∈ 1 r B such that ‖g − Tf 1‖ < 1 2 .
Now applying 6.82 with h = g − Tf 1 and ε =
4 1 , we see that
there exists f 2 ∈
2r 1 B such that ‖g − Tf 1 − Tf 2 ‖ < 1 4 .
Applying 6.82 again, this time with h = g − Tf 1 − Tf 2 and ε = 1 8
, we see that
there exists f 3 ∈
4r 1 B such that ‖g − Tf 1 − Tf 2 − Tf 3 ‖ < 1 8 .
Continue in this pattern, constructing a sequence f 1 , f 2 ,...in V. Let
∞
f = ∑ f k ,
k=1
where the infinite sum converges in V because
∞
∑
k=1
‖ f k ‖ <
∞
∑
k=1
1
2 k−1 r = 2 r ;
here we are using 6.41 (this is the place in the proof where we use the hypothesis that
V is a Banach space). The inequality displayed above shows that ‖ f ‖ < 2 r .
Because
‖g − Tf 1 − Tf 2 −···−Tf n ‖ < 1 2 n
and because T is a continuous linear map, we have g = Tf.
We have now shown that B(0, 1) ⊂ 2 r T(B). Thus 2 r B(0, 1) ⊂ T(B), completing
the proof.
Measure, Integration & Real Analysis, by Sheldon Axler