SURPRISE QUIZ I: HINTS AND SOLUTIONS
1. Let X0 ̸= {0} be a subspace of a normed linear space X. Assume for a moment that X0 is
bounded. Then there exists M > 0 such that
d(x, y) ≤ M ∀ x, y ∈ X0 . (∗)1 .
Let 0 ̸= x0 ∈ X0 . Since X0 is a subspace, αx0 ∈ X0 for all α ∈ K, and hence, (∗)1 , in
particular, implies that
|α|∥x0 ∥ = ∥αx0 ∥ = d(αx0 , 0) ≤ M ∀ α ∈ K.
Consequently,
M
|α| ≤ ∀ α ∈ K,
∥x0 ∥
a contradiction.
2. Recall from Linear Algebra that for x ∈ X, where X is a linear space
0.x = 0, (−1).x = −x.
(i) Note that ∥0∥ = ∥0.x∥ = |0|∥x∥ = 0.
(ii) From the triangle inequality, 0 = ∥0∥ = ∥x + −x∥ ≤ ∥x∥ + ∥x∥ = 2∥x∥.
(iii) Choose an α ∈ R+ ∪ {0}. We have to produce an element y ∈ X such that ∥y∥ = α.
Let 0 ̸= x ∈ X, and define y = α ∥x∥
x
. Then, y ∈ X and ∥y∥ = |α| = α.
(iv) From the triangle inequality, ∥x∥ = ∥x − y + y∥ ≤ ∥x − y∥ + ∥y∥. Therefore,
∥x∥ − ∥y∥ ≤ ∥x − y∥ ∀ x, y ∈ X. (∗)2
Interchanging the role of x and y in (∗)2 , we obtain
∥y∥ − ∥x∥ ≤ ∥x − y∥ ∀ x, y ∈ X. (∗)3
The required assertion follows from (∗)2 and (∗)3 . Note that this proves x 7→ ∥x∥ is a
uniformly continuous map.
3. We have
∥Ta (x) − Ta (y)∥ = ∥(x + a) − (y + a)∥ = ∥x − y∥ ∀ x, y ∈ X,
∥Mk (x) − Mk (y)∥ = ∥kx − ky∥ = |k|∥x − y∥ ∀ x, y ∈ X.
Therefore, for any a ∈ X and 0 ̸= k ∈ K, Ta and Mk are Lipschitz maps on the space X. In
particular, Ta and Mk are continuous maps. It is easy to see that
Ta ◦ T−a = T−a ◦ Ta = I, Mk ◦ M k1 = M k1 ◦ Mk = I,
where I is the identity map on X. These imply that Ta and Mk have inverses, namely
Ta−1 = T−a , Mk−1 = M k1 .
These inverses are also of the same type, and hence they are continuous.
4. (i) Let (zn ) be a sequence in Y1 + Y2 . Then for each n ∈ N, zn = xn + yn , where xn ∈ Y1 and
yn ∈ Y2 . Since Y1 is compact, sequence (xn ) has a subsequence (xni ) converging to a point
x ∈ Y1 . Consider the sequence (yni ) in Y2 . Again, by the compactness of Y2 , it follows that
there is a subsequence (ynir ) converging to a point y ∈ Y2 . The subsequence (xnir + ynir )
of the sequence (zn ) converges to x + y ∈ Y1 + Y2 . Thus, every sequence in Y1 + Y2 has a
convergent subsequence, proving that Y1 + Y2 is (sequentially) compact.
(ii) Consider a sequence (zn ) in Y1 + Y2 , where zn = xn + yn , xn ∈ Y1 and yn ∈ Y2 for
1
1. Let X0 ̸= {0} be a subspace of a normed linear space X. Assume for a moment that X0 is
bounded. Then there exists M > 0 such that
d(x, y) ≤ M ∀ x, y ∈ X0 . (∗)1 .
Let 0 ̸= x0 ∈ X0 . Since X0 is a subspace, αx0 ∈ X0 for all α ∈ K, and hence, (∗)1 , in
particular, implies that
|α|∥x0 ∥ = ∥αx0 ∥ = d(αx0 , 0) ≤ M ∀ α ∈ K.
Consequently,
M
|α| ≤ ∀ α ∈ K,
∥x0 ∥
a contradiction.
2. Recall from Linear Algebra that for x ∈ X, where X is a linear space
0.x = 0, (−1).x = −x.
(i) Note that ∥0∥ = ∥0.x∥ = |0|∥x∥ = 0.
(ii) From the triangle inequality, 0 = ∥0∥ = ∥x + −x∥ ≤ ∥x∥ + ∥x∥ = 2∥x∥.
(iii) Choose an α ∈ R+ ∪ {0}. We have to produce an element y ∈ X such that ∥y∥ = α.
Let 0 ̸= x ∈ X, and define y = α ∥x∥
x
. Then, y ∈ X and ∥y∥ = |α| = α.
(iv) From the triangle inequality, ∥x∥ = ∥x − y + y∥ ≤ ∥x − y∥ + ∥y∥. Therefore,
∥x∥ − ∥y∥ ≤ ∥x − y∥ ∀ x, y ∈ X. (∗)2
Interchanging the role of x and y in (∗)2 , we obtain
∥y∥ − ∥x∥ ≤ ∥x − y∥ ∀ x, y ∈ X. (∗)3
The required assertion follows from (∗)2 and (∗)3 . Note that this proves x 7→ ∥x∥ is a
uniformly continuous map.
3. We have
∥Ta (x) − Ta (y)∥ = ∥(x + a) − (y + a)∥ = ∥x − y∥ ∀ x, y ∈ X,
∥Mk (x) − Mk (y)∥ = ∥kx − ky∥ = |k|∥x − y∥ ∀ x, y ∈ X.
Therefore, for any a ∈ X and 0 ̸= k ∈ K, Ta and Mk are Lipschitz maps on the space X. In
particular, Ta and Mk are continuous maps. It is easy to see that
Ta ◦ T−a = T−a ◦ Ta = I, Mk ◦ M k1 = M k1 ◦ Mk = I,
where I is the identity map on X. These imply that Ta and Mk have inverses, namely
Ta−1 = T−a , Mk−1 = M k1 .
These inverses are also of the same type, and hence they are continuous.
4. (i) Let (zn ) be a sequence in Y1 + Y2 . Then for each n ∈ N, zn = xn + yn , where xn ∈ Y1 and
yn ∈ Y2 . Since Y1 is compact, sequence (xn ) has a subsequence (xni ) converging to a point
x ∈ Y1 . Consider the sequence (yni ) in Y2 . Again, by the compactness of Y2 , it follows that
there is a subsequence (ynir ) converging to a point y ∈ Y2 . The subsequence (xnir + ynir )
of the sequence (zn ) converges to x + y ∈ Y1 + Y2 . Thus, every sequence in Y1 + Y2 has a
convergent subsequence, proving that Y1 + Y2 is (sequentially) compact.
(ii) Consider a sequence (zn ) in Y1 + Y2 , where zn = xn + yn , xn ∈ Y1 and yn ∈ Y2 for
1
