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M382D, First Homework Solutions
Problem 1. Prove the Inverse Function Theorem in one dimension: “If
f : R → R is continuously differentiable, and if f 0 (a) 6= 0, then there
exists a neighborhood U of a and V of f (a) such that 1) f maps U in
a 1–1 manner onto V , 2) f −1 : V → U is differentiable at a, and 3)
(f −1 )0 (f (a)) = 1/f 0 (a).”
First note that being continuously differentiable is a key assumption. Otherwise the function f (x) = x + 2x2 sin(1/x) (for x 6= 0 and f (0) = 0) would
be a counterexample.
Suppose that f is not injective in a neighborhood of a. Then we can
find points xi , yi arbitrarily close to a such that f (xi ) = f (yi ). But by
Rolle’s Theorem this implies that there are points ci , between xi and yi , with
f 0 (ci ) = 0. As xi and yi approach a, so does ci , so f 0 (a) = lim f 0 (ci ) = lim 0 =
0 (where we have used the fact that f 0 is continuous at a), contradicting the
fact that f 0 (a) 6= 0.
Since f is injective and continuous, it has a continuous inverse (locally).
Now let zi be a sequence of points converging to f (a), and let xi = f −1 (zi ).
Since f and f −1 are continuous, xi → a is the same as zi → f (a). By
definition, the derivative of f −1 at f (a) is the limit of (xi − a)/(zi − f (a)) as
this happens, which is the reciprocal of the limit of (zi − f (a))/(xi − a), i.e.
1/f 0 (a). This proves parts (2) and (3).
Problem 2. Let g be a continuous real-valued function on the unit circle
in R2 such that g(0, 1) = g(1, 0) = 0 and g(−x1 , −x2 ) = −g(x1 , x2 ). Define
f : R2 → R by f (x) = |x|g(x/|x|) if x 6= 0 and f (0) = 0. (Here x ∈ R2 is
shorthand for (x1 , x2 ).)
(a) Show that f is continuous at (0, 0), that the partial derivatives of f at
(0, 0) are well-defined, and that the directional derivatives of f in every direction are well-defined.
(b) Show that f is not differentiable at (0,0) unless g is identically zero.
Since g is continuous on a compact set (namely the unit circle), it has a
maximum and minimum value. But then |f (x)| is bounded by |x| times the
maximum value of g, so limx→0 |f (x)| = 0, which implies that limx→0 f (x) =
0 = f (0), proving that f is continuous at the origin. By construction, f (x) is
identically zero on the two coordinate axes, making the partial derivatives of
f at the origin both equal to zero. (E.g. ∂f /∂x1 = lim(f (h, 0) − f (0, 0))/h =
lim 0/h = 0.) Likewise, if v is a unit vector, then the directional derivative
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in the v direction is limh→0 (f (hv) − f (0, 0))/h = limh→0 g(v) = g(v). (The
fact that g is odd implies that the limit is the same for both positive and
negative h.)
Now, if g were differentiable, then for an arbitrary unit vector v we would
have Dv f (0, 0) = v1 ∂1 f + v2 ∂2 f = 0 by the chain rule. But Dv f = g(v), so
g(v) must be zero.
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Problem 3. Let f : R2 → R be defined by f (x, y) = x|y|/ x2 + y 2 if
(x, y) 6= (0, 0) and f (0, 0) = 0. (This is a function of the kind considered in
Problem 3.) Compute the partial derivatives of f on a neighborhood of the
origin, and show that they are well-defined everywhere but are not continuous
at the origin.
As noted in my email, this is incorrect as stated, and ∂f /∂y is NOT
well-defined
at y = 0. However, at all points where y 6= 0, we have f (x, y) =
√
±xy/ x2 + y 2 . Taking derivatives by formula, I get that ∂f /∂x = |y 3 |/(x2 +
y 2 )3/2 and that ∂f /∂y = ±x3 /(x2 + y 2 )3/2 , with the ± being the sign of y.
In polar coordinates, this is ∂f /∂x = | sin3 (θ)| and ∂f /∂y = ± cos3 (θ). If
we approach the origin along any non-horizontal line (i.e. where both are
well-defined), then the limit of at least one of these is non-zero, and hence is
NOT equal to the partial derivatives AT the origin.
Problem 4. Let x ∈ R. A derivation at x is a map D : C ∞ (R) → R such
that, for any smooth functions f, g and any scalar c,
D(f + g) = D(f ) + D(g)
D(cf ) = cD(f )
D(f g) = f (x)D(g) + g(x)D(f )
Let Vx be the set of derivations at x.
(a) Show that d/dx, followed by evaluation at a point x, is a derivation at x.
(b) Show that Vx is a vector space, with the obvious notions of addition and
scalar multiplication.
(c) Show that Vx is one-dimensional, with basis d/dx. (Hint: Use Taylor’s
theorem).
(d) Let V be the disjoint union of all the sets Vx . Define a reasonable topology
for V such that V is connected.
Part (a) is just the basic rules of calculus: (f + g)0 (x) = f 0 (x) + g 0 (x),
(cf )0 (x) = cf 0 (x) and (f g)0 (x) = f (x)g 0 (x) + g(x)f 0 (x).
(b) Defining (D1 + D2 )(f ) = D1 (f ) + D2 (f ) and (cD1 )(f ) = cD1 (f ), it’s easy
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to see that D1 + D2 and cD1 are derivations. Since the set of derivations is
closed under scalar multiplication and addition, it’s a vector space.
(c) Let D be a derivation, and let k be D applied to the function f (x) = x.
(For this part, I’m going to call the point where we’re evaluating a instead
of x, so that x can be a variable.) Now we check that D(1) = D(12 ) =
1D(1) + 1D(1) = 2D(1) by the product rule, so D(1) = 0. Meanwhile,
D(x − a) = k. If g is any smooth function, then D((x − a)g(x)) = (a −
a)D(g) + g(a)D(x − a) = kg(a). Now let f (x) by any smooth function. We
can always write f (x) = f (a) + (x − a)g(x) for some smooth function g(x)
with g(a) = f 0 (a). (Why is g(x) smooth? Because if the n-th derivative
of g didn’t exist then the (n + 1)st derivative of f wouldn’t exist.)But then
Df = 0 + kf 0 (a) + 0 = kf 0 (a). Since this is true for arbitrary f , we must
have D = kd/dx (evaluated at a).
(d) Since every differential is of the form “kd/dx evaluated at a”, we can
identify V with R2 = {(a, k)} and use the obvious topology on R2 . This
means that a sequence of derivations Di converges to a derivation D if an
only if (i) the points where they are acting on converge, and (ii) the values
of Di applied to the function f (x) = x converge.
Problem 5. Now generalize to Rn . A derivation at x is, as before, a linear
map D : C ∞ (Rn ) → R that satisfies the above axioms. Show that Vx is an
n-dimensional vector space with basis {∂/∂xi }.
The only difference is in the Taylor expansion. We need to show that
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f (x) = f (a)+ (xi −ai )gi (x) for some set of smooth functions gi (x), and with
gi (a) = ∂i f (a). But that follows immediately from the following expansion,
which I’ll spell out for n = 3 but whose generalization is clear: f (x1 , x2 , x3 ) =
f (a1 , a2 , a3 ) + (f (x1 , a2 , a3 ) − f (a1 , a2 , a3 )) + (f (x1 , x2 , a3 ) − f (x1 , a2 , a3 )) +
(f (x1 , x2 , x3 ) − f (x1 , x2 , a3 )) and 1-dimensional calculus applied to each of
the three differences. Define ki = D applied to the function (xi − ai ). I claim
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P
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that D = ki ∂i , since Df (x) = D(f (a)) + gi (a)D(xi − ai ) = ki ∂i f (a).
The upshot of problems 4 and 5 is that the tangent space to Rn can
be defined purely algebraically from the ring of smooth functions on Rn .
Since this is a local operation, the tangent space of an arbitrary smooth
manifold can be similarly constructed from the ring of smooth functions on
that manifold. That’s not an angle we will be emphasizing in this class, but
it’s at the heart of non-commutative geometry, and also of algebraic geometry
(with “smooth” replaced by “analytic” or “algebraic”).
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Problem 6: Let A0 be an m × n matrix of maximal rank. Show that there
is a neighborhood U of A0 in the space of m × n matrices (i.e. of Rnm )
and an algorithm for finding a basis for the kernel (for m ≤ n) and image
(for m ≥ n) of each A ∈ U such that these bases vary continuously with
the entries of A. In other words, not only is the condition “A is of maximal
rank” stable, but the kernels and images don’t change much as we vary A.
First note that if A0 has maximal rank, then there is an m × m or n × n
sub-matrix of A0 whose determinant is non-zero. The determinant of the
analogous submatrix of A is a smooth (actually polynomial) function of the
entries of A, and hence is a continuous function on Rnm . Since it is nonzero
at A0 , it is nonzero on a neighborhood of A0 , implying that every matrix in
a neighborhood of A0 has maximal rank.
Now suppose that m ≥ n. Then all n columns of A are linearly independent, and these columns form a basis for the column space of A. (That was
easy!)
Now suppose that m ≤ n. Then there are m linearly independent
columns. Suppose that they are the first m columns.(Otherwise re-order
the variables) Write A = (A1 |A2 ), where A1 is an invertible m × m matrix
and A2 is m × (n − m). A then row-reduces to (I|A−1
1 A2 ). Let bi be a vector
whose first m entries are minus the i-th column of A−1
1 A2 , whose m + i th
entry is 1, and whose other entries are 0. The bi ’s give a basis for the null
space of A, and their entries vary continuously with the entries of A.
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