Integration: Exercises R Ch 5
Last updates: 9 April 2011
Integration: Exercises R Ch 5
- Let consist of two points and ,
, and let
be the resulting real -space. Identify
each real function on with the point
in the plane, and sketch the unit balls of
. Note that they are
convex if and only if .
For which is this unit ball a square? A circle? If
, how does this situation differ from the preceding one?
- Prove that the unit ball (open or closed) is convex in every normed linear space.
prove that the unit ball of
is strictly convex; this means that if
. (Geometrically, the surface of the ball contains no straight lines.)
Show that this fails in every ,
in every ,
and in every . (Ignore trivialities, such as
spaces consisting of only one point.)
- Let be the space of all continuous functions on
, with the supremum norm. Let
consist of all for which
Prove that is a closed convex subset of which
contains no element of minimal norm.
- Let be the set of all ,
relative to Lebesgue measure, such that
Show that is a closed convex subset of
which contains infinitely many elements of minimal norm. (Compare this and Exercise 4 with Theorem 4.10.)
Let be a bounded linear functional on a subspace
of a Hilbert space . Prove that has a unique
norm-preserving extension to a bounded linear functional on , and that this extension
vanishes on .
- Construct a bounded linear functional on some subspace of
which has two (hence infinitely many) distinct norm-preserving linear extensions to
- Let be a normed linear space, and let
be its dual space with the norm
Prove that is a Banach space.
Prove that the mapping
is, for each , a bounded linear functional on
, of norm
. (This gives a natural imbedding of
in its "second dual" ,
the dual space of .)
Prove that is
is a sequence in such
is bounded for every .
- Let , ,
and be the Banach spaces consisting of all
complex sequences ,
defined as follows:
is the subspace of
consisting of all for which
if and only if
if and only if
Prove the following four statements.
is a bounded linear functional on
is obtained this way. In brief,
In the same sense, .
Every induces a
bounded linear functional on , as in (a).
However, this does not give all of
contains nontrivial functionals that vanish on all of
are separable but is not.
If converges for every sequence
as , prove that
- For ,
let denote the space
of all complex functions
on for which
Prove that is a Banach space, if
; also if
(The members of are said to satisfy a Lipschitz
condition of order .)
- Let be a triangle (two dimensional figure) in the plane, let
be the set consisting of the vertices of , and let be
the set of all real functions on , of the form
Show that to each
there corresponds a unique measure on
(Compare Sec. 5.22.)
Replace by a square, let again be the set of its
vertices, and let be as above. Show that to each point of
there still corresponds a measure on , with the above property, but that
uniqueness is now lost.
Can you extrapolate to a more general theorem? (Think of other figures, higher dimensional spaces.)
- Let be a sequence
of continuous complex functions on a (nonempty) complete metric space ,
such that exists (as a complex number) for every
Prove that there is an open set and
an number such that
and for .
prove that there is an open set and
an integer such that
if and .
- Hint for (b): For ,
some has a nonempty interior.
- Let be the space of all real continuous functions on
with the supremum norm.
Let be the subset of
consisting of those for which there exists a
for all . Fix
and prove that each open set in contains an open set which does not intersect
can be uniformly approximated by a zigzag function with very large
slopes, and if is small,
.) Show that this implies the
existence of a dense in
which consists entirely of nowhere differentiable functions.
be an infinite matrix with complex entries, where .
associates with each sequence
, defined by
provided that these series converge.
Prove that transforms every convergent sequence
to a sequence
which converges to the
same limit if and only if the following conditions are satisfied:
The process of passing from
is called a
summability method. Two examples are
for each .
Prove that each of these also transforms some divergent sequences
(even some unbounded ones) to convergent sequences
- Suppose and are Banach spaces, and suppose
is a linear mapping of into ,
with the following property: For every sequence
in for which
exist, if is true that . Prove that
This is the so called "closed graph theorem". Hint: Let
be the set of all ordered pairs ,
with addition and scalar multiplication defined componentwise. Prove that
is a Banach space, if . The graph of is the subset of
formed by the pairs ,
. Note that our hypothesis says that
is closed; hence is a Banach space. Note that
is continuous, one-to-one, and linear and maps
Observe that there exist nonlinear mappings (of
onto , for instance) whose graph is closed although
they are not continuous: if ,
- If is a ositive measure, each
a multiplication operator
. Prove that
For which measures is it true that
is a sequence of bounded linear transformations from a normed linear space
to a Banach space , suppose
for all ,
and suppose there is a dense set such that
converges for each . Prove that
converges for each .
- If is the th partial
sum of the Fourier series of a function
prove that uniformly,
as , for each . That is, prove that
On the other hand, if , prove that there exists an
that the sequence is unbounded. Hint:
Apply the reasoning of Exercise 18 and that of Sec. 511, with a better estimate of
than was used there.
- (a) Does there exist a sequence of continuous positive functions
on such that
unbounded if and only if is rational?
Replace "rational" by "irrational" in (a) and answer the resulting question.
as " and answer the resulting analogues of (a) and (b).
- Suppose is
measurable, and .
Must there be a translate of
that does not intersect ? Must there be a homeomorphism
does not intersect ?
and for some
. (See Exercise 11.) Prove that the
Fourier series of converges to ,
by completing the following outline: It is enough to consider the case
, . THe difference between the partial sums
tends to 0 as . The function
. Apply the
Riemann-Lebesgue lemma. More careful reasoning shows that the convergence is actually uniform
Notes and References
These exercises are taken from [Ru, Chapt. 5] for a course in "Measure Theory" at the Masters level at University of Melbourne.
Principles of Mathematical Analysis, Third edition, McGraw-Hill, 1976.
Real and complex analysis, Third edition, McGraw-Hill, 1987.