8.4. EXERCISES 215

9. If U,V are open sets in Banach spaces X ,Y respectively and f : U →V is one to oneand onto and both f , f−1 are C1, show that D f (x) : X →Y is one to one and onto foreach x ∈U . Hint: f ◦ f−1 = identity. Now use chain rule.

10. A function f : U ⊆C→ C where U is an open set subset of the complex numbers Cis called analytic if

limh→0

f (z+h)− f (z)h

≡ f ′ (z) , z = x+ iy

exists and z→ f ′ (z) is continuous. Show that if f is analytic on an open set U and iff ′ (z) ̸= 0, then there is an open set V containing z such that f (V ) is open, f is one toone, and f , f−1 are both continuous. Hint: This follows very easily from the inversefunction theorem. Recall that we have allowed for the field of scalars the complexnumbers.

11. Problem 8 has to do with concluding that f is locally one to one if D f (x0) is onlyknown to be one to one. The next obvious question concerns the situation whereD f (x0) maybe is possibly not one to one but is onto. There are two parts, a linearalgebra consideration, followed by an application of the inverse function theorem.Thus these two problems are each generalizations of the inverse function theorem.

(a) Suppose X is a finite dimensional vector space and M ∈ L (X ,Y ) is onto Y .Consider a basis for M (X) = Y,{Mx1, · · · ,Mxn} . Verify that {x1, · · · ,xn} islinearly independent. Define X̂ ≡ span(x1, · · · ,xn). Show that if M̂ is the re-striction of M to X̂ , then M is one to one and onto Y .

(b) Now suppose f : U ⊆ X → Y is C1 and D f (x0) is onto Y . Show that there is aball B( f (x0) ,ε) and an open set V ⊆ X such that f (V ) ⊇ B( f (x0) ,ε) so thatif D f (x) is onto for each x ∈U , then f (U) is an open set. This is called theopen map theorem. You might use the inverse function theorem with the spacesX̂ ,Y . You might want to consider Problem 1. This is a nice illustration of whywe developed the inverse and implicit function theorems on arbitrary normedlinear spaces. You will see that this is a fairly easy problem.

12. Recall that a function f : U ⊆ X → Y where here assume X is finite dimensional, isGateaux differentiable if

limt→0

f (x+ tv)− f (x)t

≡ Dv f (x)

exists. Here t ∈ R. Suppose that x→ Dv f (x) exists and is continuous on U . Showit follows that f is differentiable and in fact Dv f (x) = D f (x)v. Hint: Let g(y) ≡f (∑i yixi) and argue that the partial derivatives of g all exist and are continuous.Conclude that g is C1 and then argue that f is just the composition of g with a linearmap.

13. Let

f (x,y) =

 (x2−y4)2

(x2+y4)2 if (x,y) ̸= (0,0)

1 if (x,y) = (0,0)

Show that f is not differentiable, and in fact is not even continuous, but Dv f (0,0)exists and equals 0 for every v ̸= 0.