SOLUTIONS FOR THE MIDTERM

MATH 25

1. An application of Taylor’s expansion

In this problem, a ∈ R and f : R → R is a C^{∞} function. The goal of the problem is to prove that x ↦ (f(x) − f(a))/(x − a), which is defined on R \ {a}, extends to a C^{∞} function on R.

1.1. 8pts. Prove that there exists a continuous map g : R → R such that for every x ≠ a, we have

g(x) = \frac{f(x) − f(a)}{x − a}.

What is g(a) equal to?

Solution: The function g is continuous on R \ {a}. If we set g(a) = f′(a), then the fact that f is differentiable at x = a implies that lim_{x→a}(f(x) − f(a))/(x − a) = f′(a) = g(a) so that g is continuous at x = a. This way, g is continuous on R and in addition, g(a) = f′(a).

1.2. 8pts. Prove that g : R → R is C^{1} and find g′(a).

Solution: Taylor’s expansion around x = a tells us that:

f(x) = f(a) + (x − a)f′(a) + (x − a)^{2} \frac{f′′(a)}{2} + \frac{(x − a)^{3}}{2} ∫_{0}^{1}(1 − u)^{2}f′′′(a + u(x − a))du

and therefore that

g(x) = f′(a) + (x − a)\frac{f′′(a)}{2} + \frac{(x − a)^{2}}{2} ∫_{0}^{1}(1 − u)^{2}f′′′(a + u(x − a))du

so that

\frac{g(x) − g(a)}{x − a} = \frac{f′′(a)}{2} + \frac{x − a}{2} ∫_{0}^{1}(1 − u)^{2}f′′′(a + u(x − a))du

from which we deduce that lim_{x→a}(g(x) − g(a))/(x − a) = f′′(a)/2. This proves that g′(a) = f′′(a)/2.

To prove that g is C^{1} we could invoke the above formulas and “differentiability under the integral” but let us do it “by hand”. By applying the quotient rule, we see that g is C^{1} on R \ {a} so in order to prove that g is C^{1}, we need to show that g′(x) → g′(a) as x → a. The quotient rule tells us that if x ≠ a, then

g′(x) = \frac{(x − a)f′(x) − (f(x) − f(a))}{(x − a)^{2}} = \frac{f′(x) − f′(a)}{x − a} + \frac{f′(a) − g(x)}{x − a}.

When x → a, the first term above converges to f′′(a). The second term is equal to:

−\frac{f′′(a)}{2} − \frac{x − a}{2} ∫_{0}^{1}(1 − u)^{2}f′′′(a + u(x − a))du

and therefore converges to −f′′(a)/2 as x → a. This proves that g′(x) → g′(a) as x → a and therefore that g′(x) is continuous at x = a.

1.3. 12pts. Assume for this question that f(a) = f′(a) = ⋯ = f^{(n+1)}(a) = 0, where n ∈ Z_{≥0}. Prove that g : R → R is C^{n} and that g(a) = g′(a) = ⋯ = g^{(n)}(a) = 0.

Solution: In this case, Taylor’s expansion is just:

f(x) = \frac{(x − a)^{n+2}}{(n + 1)!} ∫_{0}^{1}(1 − u)^{n+1}f^{(n+2)}(a + u(x − a))du,

and therefore

g(x) = \frac{(x − a)^{n+1}}{(n + 1)!} ∫_{0}^{1}(1 − u)^{n+1}f^{(n+2)}(a + u(x − a))du.

Since g(x) is the product of (x − a)^{n+1} by the function

\frac{1}{(n + 1)!} ∫_{0}^{1}(1 − u)^{n+1}f^{(n+2)}(a + u(x − a))du

which is C^{∞} as can be seen by differentiating under the integral sign for example, we see that g is C^{n} (actually C^{∞}) and that g(a) = g′(a) = ⋯ = g^{(n)}(a) = 0.

1.4. 12pts. Return now to the case of a general C^{∞} function f : R → R. Prove that g : R → R is C^{∞} and compute g^{(n)}(a) for n ≥ 0.

Solution: We’ve already seen that g is C^{∞}; choose n ≥ 0 and set

f_{0}(x) = f(x) − f(a) − (x − a)f′(a) − ⋯ − \frac{(x − a)^{n+1}}{(n + 1)!} f^{(n+1)}(a)

so that f_{0}(a) = f′_{0}(a) = ⋯ = f_{0}^{(n+1)}(a) = 0. We have

g_{0}(x) = g(x) − f′(a) − ⋯ − \frac{(x − a)^{n}}{(n + 1)!} f^{(n+1)}(a)

and by the previous question, we know that g_{0}(a) = g′_{0}(a) = ⋯ = g_{0}^{(n)}(a) = 0. We see that g_{0}^{(n)}(a) = g^{(n)}(a) − \frac{n!}{(n+1)!}f^{(n+1)}(a) so that g^{(n)}(a) = f^{(n+1)}(a)/(n + 1).

2. Derivations of C^{∞}(R, R)

In this problem, we will use the results from problem 1. Let C^{∞}(R, R) denote the set of all maps f : R → R which are C^{∞}. The goal of this problem is to find all maps T : C^{∞}(R, R) → C^{∞}(R, R) satisfying the following three properties:

(1) T (f + g) = T (f) + T (g);

(2) T (λf) = λT (f) if λ ∈ R;

(3) T (fg) = fT (g) + gT (f). Observe that T (φ) = φ′ is one such map.

2.1. 8pts. Let 1 denote the constant function x ↦ 1. What is T (1)?

Solution: We have T (1) = T (1 · 1) = 1T (1) + 1T (1) = 2T (1) so T (1) = 0.

2.2. 8pts. For a ∈ R, let θ_{a} : R → R be the map x ↦ (x − a)^{2} and let ε_{a} = T (θ_{a}). What is ε_{a}(a)?

Solution: We have T ((x − a)^{2}) = 2(x − a)T (x − a) so that ε_{a}(a) = 0.

2.3. 12pts. Let φ ∈ C^{∞}(R, R) be a function such that φ′(a) = 0. Prove that T (φ)(a) = 0.

By the preceding problem, we know that the function ψ(x) = (φ(x) − φ(a))/(x − a) is C^{∞} and that ψ(a) = 0. So we apply the preceding problem once again and deduce that χ(x) = (φ(x)−φ(a))/(x−a)^{2} is C^{∞} so that we can write φ(x) = φ(a)+θ_{a}(x)χ(x). Therefore we have T (φ) = T (θ_{a})χ + θ_{a}T (χ) and since T (θ_{a})(a) = θ_{a}(a) = 0, we have T (φ)(a) = 0.

2.4. 12pts. Prove that there exists f ∈ C^{∞}(R, R) such that for all φ ∈ C^{∞}(R, R), we have T (φ) = fφ′.

Solution: observe that T (x − a) does not depend on a since T (a) = 0. Set f = T (x − a). The function φ(x) − φ′(a)(x − a) has its first derivative = 0 when x = a so that

T [φ(x) − φ′(a)(x − a)](a) = 0

by the preceding question. This implies that T (φ)(a) = φ′(a)f(a). This is true for all a ∈ R, so we have T (φ) = fφ′.

3. A characterization of ruled functions

Let I = [a, b] be a closed interval, and let f : I → R be a function.

The goal of this problem is to prove that a function f : I → R is ruled if and only if at every x ∈ I, f has a left limit and a right limit (at x = a we only ask for a right limit and at x = b we only ask for a left limit).

Let L(I) be the set of functions f : I → R such that at every x ∈ I, f has a left limit and a right limit (at x = a we only ask for a right limit and at x = b we only ask for a left limit). Let S(I) be the set of step functions f : I → R. If f : I → R is a bounded function, let |f|_{I} denote sup_{x∈I} |f(x)|.

3.1. 12pts. Prove that if f ∈ L(I), then f is bounded.

Solution: if this was not the case, there would exist a sequence {x_{n}}_{n≥1} of elements of I such that f(x_{n}) → +∞. Since I is compact, we can extract from x_{n} a subsequence x_{φ(n)} converging to some x ∈ I. By extracting again, we can assume that either x_{φ∘ψ(n)} > x or x_{φ∘ψ(n)} < x for all n. Let us assume that the first case holds. Since x_{φ∘ψ(n)} > x and x_{φ∘ψ(n)} → x, we have f(x_{φ∘ψ(n)}) → y^{+} where y^{+} is the right limit of f at x. This contradicts the fact that the f(x_{φ∘ψ(n)}) are unbounded. Therefore, if f ∈ L(I), then f is bounded.

3.2. 12pts. Prove that if one defines d(f, g) = |f − g|_{I} for f, g ∈ L(I), then (L(I), d) is a complete metric space.

Solution: It is easy to check that d satisfies the three requirements to be a distance on L(I), so we will now prove that (L(I), d) is complete. We have to prove that if {f_{n}}_{n≥1} is a sequence of elements of L(I) which converges uniformly to a function f, then f has a right limit and a left limit at each x ∈ I. Let us do the case of a left limit, the case of a right limit being analogous. Choose x ∈ I and a sequence {x_{n}}_{n≥1} such that x_{n} < x and x_{n} → x. Choose ε > 0. There exists m ≥ 0 such that |f − f_{m}|_{I} < ε. Since f_{m}(x_{n}) → y^{−} as n → ∞, there exists N ≥ 0 such that if k, ℓ ≥ N then |f_{m}(x_{k}) − f_{m}(x_{ℓ})| < ε. Now we have

|f(x_{k}) − f(x_{ℓ})| < |f(x_{k}) − f_{m}(x_{k})| + |f_{m}(x_{k}) − f_{m}(x_{ℓ})| + |f_{m}(x_{ℓ}) − f(x_{ℓ})|,

so that |f(x_{k}) − f(x_{ℓ})| < 3ε if k, ℓ ≥ N. This proves that for every sequence {x_{n}}_{n≥1} such that x_{n} < x and x_{n} → x, f(x_{n}) has a limit. By “intertwining” sequences, it is clear that this limit is unique and therefore that f has a left limit at x. This proves that f ∈ L(I) and therefore that (L(I), d) is a complete metric space.

3.3. 12pts. Explain why S(I) ⊂ L(I). Prove that (S(I), d) is dense in (L(I), d). Hint: prove that if f ∈ L(I) and ε > 0, then the set of x ∈ I such that there exists φ ∈ S(I) satisfying sup_{y∈[a,x]} |f(y) − φ(y)| ≤ ε is both open and closed.

Solution: S(I) ⊂ L(I) because a step function has a left limit and a right limit at every point (this is clear from the definition). Let us now prove that (S(I), d) is dense in (L(I), d). We will first show that this follows from the hint. Choose ε > 0; if the set of x ∈ I such that there exists φ ∈ S(I) satisfying sup_{y∈[a,x]} |f(y) − φ(y)| ≤ ε is both open and closed, then it is a connected component of I and it has to be empty or I itself since I is connected. Since this set contains a at least, we see that it is I itself. By the definition of this set, there exists φ ∈ S(I) satisfying |f(y) − φ(y)|_{I} ≤ ε. This shows that f is a uniform limit of elements of S(I).

Now, we will prove the hint. Let A(f, ε) be the set of x ∈ I such that there exists φ ∈ S(I) satisfying sup_{y∈[a,x]} |f(y) − φ(y)| ≤ ε. If x ∈ A(f, ε), then obviously any x′ ≤ x is in A(f, ε) also; this implies that A(f, ε) = [a, x) or that A(f, ε) = [a, x] for some x. We will first show that A(f, ε) is open, so suppose that A(f, ε) = [a, x]. If x = b, we’re done. Otherwise, f has a right limit y^{+} at x. We claim that this implies that there exists δ > 0 such that if x < z < x + δ then |f(z) − y^{+}| < ε/2 . If this was not the case, we could find a sequence z_{n} violating the claim and converging to x, which would contradict the fact that f has a right limit at x. The claim is therefore true, and if we have φ such that sup_{y∈[a,x]} |f(y)−φ(y)| ≤ ε, then we can extend φ to [a, x + δ) by setting φ = y^{+} on (x, x + δ) and φ(x) = f(x). We now prove that A(f, ε) is closed. Suppose that there exists x and some sequence x_{n} → x and step functions φ_{n} on [a, x_{n}] such that sup_{y∈[a,x_{n}]} |f(y) − φ_{n}(y)| ≤ ε. Since f has a left limit y^{−} at x, an argument similar to the one above shows that there exists δ > 0 such that if x − δ < z < x then |f(z) − y^{−}| < ε/2 and if we choose n such that x_{n} > x − δ, then the step function φ defined by φ = φ_{n} on [a, x_{n}] and φ = y^{−} on (x_{n}, x) and φ = f(x) at x is within ε of f and therefore, x ∈ A(f, ε). This proves the hint.

3.4. 4pts. Conclude: prove that L(I) is equal to the set of ruled functions on I.

Solution: a function is ruled if and only if it is a uniform limit of elements of S(I) and so if and only if it is in L(I) so that L(I) coincides with the set of all ruled functions.