The Inverse Function Theorem and Introduction to Infinite Series Flashcards

Question 1

Q

What is a strictly increasing function

Answer

A

Let A ⊆ R. A function f : A → R is strictly
increasing, if x1, x2 ∈ A, x1 < x2 implies f(x1) < f(x2)

Question 2

Q

What is the Inverse Function Theorem

Answer

A

Assume that
the function f is -
* Defined on a closed interval [a, b];
* Strictly increasing;
* Continuous on [a, b].

Then there exists a function g which is -
* Defined on the closed interval [f(a), f(b)];
* Strictly increasing;
* Continuous on [f(a), f(b)],
and which is inverse to f, that is, g(f(x)) = x for all x ∈ [a, b]

Question 3

Q

Explain the proof of the Existence of the inverse on [f(a), f(b)]

Answer

A

First change the codomain of f from R to [f(a), f(b)]. We note that for any x ∈ (a, b) we have a < x < b. Since f is strictly increasing, f(a) < f(x) < f(b) and so f(x) ∈
(f(a), f(b)). If x = a or x = b then it is also true that f(x) ∈ [f(a), f(b)]. Therefore, f has domain [a, b] and codomain [f(a), f(b)].
To prove surjectivity. Take any d ∈ [f(a), f(b)]. Since f
is continuous, by the Intermediate Value Theorem there exists c ∈ [a, b] such that f(c) = d.
To prove injectivity. Take x1, x2 ∈ [a, b] such that x1 ≠ x2. Since f is strictly increasing, if x1 < x2, then f(x1) < f(x2), and if x1 > x2, then f(x1) > f(x2). f(x1) ≠ f(x2), which means that f is injective
Therefore, f : [a, b] → [f(a), f(b)] is bijective and so By there exists g which is the inverse of f

Question 4

Q

Explain the proof that g is strictly increasing

Answer

A

Assume not, then there exist y1, y2 ∈ [f(a), f(b)] such that y1 < y2 but g(y2) ≤ g(y1).
We cannot have g(y2) = g(y1) because g is injective. So g(y2) < g(y1).
Since f is strictly increasing, f(g(y2)) < f(g(y1)). This would lead to y2 < y1, contradicting y1 < y2

Question 5

Q

Explain the proof that g is continuous on [f(a), f(b)] (Won’t ask)

Answer

A

We take an arbitrary y ∈ [a, b] and show that g is continuous at y. We denote x = g(y).
Let ε > 0. To verify the definition of “continuous at y”, we need to find δ > 0 such that
whenever z ∈ [f(a), f(b)] and y − δ < z < y + δ, one has x − ε < g(z) < x + ε.
We first find δ1 > 0 such that z ∈ [f(a), f(b)] and z < y + δ1 =⇒ g(z) < x + ε.
If x+ε ≤ b, then put y+δ1 = f(x+ε). This guarantees δ1 > 0 since f is strictly increasing, and so f(x) = y < f(x + ε). If z < y + δ1, then g(z) < g(y + δ1) = g(f(x + ε)) = x + ε.
If, however, x + ε > b, we can choose δ1 however we want: e.g., put δ1 = 1000. Indeed, if z ∈ [f(a), f(b)] then g(z) ≤ b and so automatically g(z) < x + ε.
We now find δ2 > 0 such that z ∈ [f(a), f(b)] and y − δ2 < z =⇒ x − ε < g(z).
Finding δ2 is completely similar to finding δ1, and we skip the details.
It remains to take δ = min{δ2, δ1}. Now, if z ∈ [f(a), f(b)] and y − δ < z < y + δ, then z satisfies both (2) and (1), and so x − ε < g(z) and g(z) < x + ε, as required.