0.3. The geometry of Hilbert spaces 25
an orthogonal basis uj, 1 ≤ j ≤ n, such that
αjuj. Then by the triangle and Cauchy–Schwarz inequalities,
and we can choose m2 =
In particular, if fn is convergent with respect to . 2, it is also convergent
with respect to . 1. Thus .
is continuous with respect to .
its minimum m 0 on the unit sphere (which is compact by the Heine–Borel
theorem, Theorem 0.12). Now choose m1 = 1/m.
Problem 0.18. Show that the norm in a Hilbert space satisfies f + g =
f + g if and only if f = αg, α ≥ 0, or g = 0.
Problem 0.19 (Generalized parallelogram law). Show that, in a Hilbert
xj − xk
The case n = 2 is (0.47).
Problem 0.20. Show that the maximum norm on C[0, 1] does not satisfy
the parallelogram law.
Problem 0.21. In a Banach space, the unit ball is convex by the triangle
inequality. A Banach space X is called uniformly convex if for every
ε 0 there is some δ such that x ≤ 1, y ≤ 1, and
≥ 1 − δ imply
x − y ≤ ε.
Geometrically this implies that if the average of two vectors inside the
closed unit ball is close to the boundary, then they must be close to each
Show that a Hilbert space is uniformly convex and that one can choose
δ(ε) = 1 − 1 −
. Draw the unit ball for
for the norms x
|x1| + |x2|, x
= |x1|2 + |x2|2, and x
= max(|x1|, |x2|). Which of
these norms makes
(Hint: For the first part, use the parallelogram law.)
Problem 0.22. Suppose Q is a vector space. Let s(f, g) be a sesquilinear
form on Q and q(f) = s(f, f) the associated quadratic form. Prove the
(0.54) q(f + g) + q(f − g) = 2q(f) + 2q(g)