- Suppose L ∈ℒ where
X and Y are two finite dimensional normed vector
spaces and suppose L is one to one. Show there exists r > 0 such that for all
x ∈ X,
Hint: Show that
≡ is a norm. Now suppose
L ∈ℒ is one to one
and onto for
X,Y Banach spaces. Explain why the same result holds. Hint: Recall
open mapping theorem.
- Suppose B is an open ball in X, a Banach space, and f : B → Y is differentiable.
Suppose also there exists L ∈ℒ such that
for all x ∈ B. Show that if x1,x2 ∈ B,
Hint: Consider Tx = f
− Lx and argue
- ↑ Let U be an open subset of X,f : U → Y where X,Y are finite dimensional
normed linear spaces and suppose f ∈ C1 and
Df is one to one. Then show
f is one to one near x0. Hint: Show using the assumption that f is C1 that there
exists δ > 0 such that if
then use Problem 1. In case X,Y are Banach spaces, assume Df is one to one
- Suppose U ⊆ X is an open subset of X a Banach space and that f : U → Y is
differentiable at x0 ∈ U such that Df is one to one and onto from
X to Y .
−1 ∈ℒ) Then show that
≠f for all
x sufficiently near
but not equal to x0. In this case, you only know the derivative exists at
- Suppose M ∈ℒ where
X and Y are finite dimensional linear spaces and
suppose M is onto. Show there exists L ∈ℒ such that
where P ∈ℒ, and
P2 = P. Also show L is one to one and onto from X1
to Y. Hint: Let be a basis of
Y and let Mxi = yi. Then
,xn} is a linearly independent set and show you can obtain
,xm}, a basis for X in which Mxj = 0 for j > n. Then
- ↑ Let f : U ⊆ X → Y,f is C1, and Df is onto for each
x ∈ U. Then show f maps
open subsets of U onto open sets in Y . Hint: Let P = LDf as in Problem
Argue L maps open sets from Y to open sets of X1 ≡ PX and L−1 maps open sets
from X1 to open sets of Y. Then Lf =
v + o
. Now for
z ∈ X1, let h =
. Then h is C1 on some small open subset of
X1 containing 0 and Dh =
LDf which is seen to be one to one and
onto and in
. Therefore, if r is small enough, h equals
an open set in
X1, V. This is by the inverse function theorem. Hence
V and so f
− f =
, an open
set in Y.
- Suppose U ⊆ ℝ2 is an open set and f : U → ℝ3 is C1. Suppose Df has rank
Show that for near
, the points
f may be realized in one of the
This shows that parametrically defined surfaces can be obtained locally in a
particularly simple form.
- Let f : U → Y , Df exists for all
x ∈ U, B
⊆ U, and there exists
L ∈ℒ, such that
L−1 ∈ℒ, and for all
x ∈ B
Show that there exists ε > 0 and an open subset of B
,V , such that
f : V → B is one to one and onto. Also
Df−1 exists for each
y ∈ B and is given by the formula
< , consider
}. This is a version of the inverse
function theorem for f only differentiable, not C1.
- Denote by C the space of functions which are continuous having values
X and define a norm on this linear space as follows.
Show for each λ ∈ ℝ, this is a norm and that C is a complete normed
linear space with this norm.
- ↑Let f :
× X → X be continuous and suppose f satisfies a Lipschitz
and let x0 ∈ X. Show there exists a unique solution to the Cauchy problem,
for t ∈.
Hint: Consider the map
where the integral is defined componentwise. Show G is a contraction map for
given in Problem 9 for a suitable choice of λ and that therefore, it has a unique
fixed point in C. Next argue, using the fundamental theorem
of calculus, that this fixed point is the unique solution to the Cauchy
- ↑Use Theorem 6.7.5 to give another proof of the above theorem. Hint: Use the
same mapping and show that a large power is a contraction map.
- Suppose you know that u
≤ a + ∫
ds where k
≥ 0 and k is in
. Show that then u
. This is a version of
Gronwall’s inequality. Hint: Let W =
ds. Then explain why
≤ ak. Now use the usual technique of an integrating factor
you saw in beginning differential equations.
- ↑Use the above Gronwall’s inequality to establish a result of continuous dependence
on the initial condition and f in the ordinary differential equation of Problem
- The existence of partial derivatives does not imply continuity as was shown in an
example. However, much more can be said than this. Consider
Show the directional derivative of f at exists and equals 0 for every
direction. The directional derivative in the direction
Now consider the curve x2 = y4 and the curve y = 0 to verify the function fails to
be continuous at .
Show that this function is not continuous at but that it has all directional
and they all equal 0.
- Let Xi be a normed linear space having norm
i. Then we can make
i=1nXi into a normed linear space by defining a norm on x ∈∏
Show this is a norm on ∏
i=1nXi as claimed.
- Suppose f : U ⊆ X × Y → Z and D2f
−1 ∈ℒ exists and
f is C1 so
the conditions of the implicit function theorem are satisfied. Also suppose that all
these are complex Banach spaces. Show that then the implicitly defined function
y = y is analytic. Thus it has infinitely many derivatives and can be given as a
power series as described above.