12 Lecture 3

Vector Spaces of Functions

3.1 of Continuous Functions C(I)

Let I ⊆ R be an . Then I is of the form (for some a < b)   {x ∈ R| a < x < b}, an open interval;  {x ∈ R| a ≤ x ≤ b}, a closed interval; I =  {x ∈ R| a ≤ x < b}  {x ∈ R| a < x ≤ b}.

Recall that the space of all functions f : I −→ R is a . We will now list some important subspaces: Example (1). Let C(I) be the space of all continuous functions on I. If f and g are continuous, so are the functions f + g and rf (r ∈ R). Hence C(I) is a vector space. Recall, that a is continuous, if the graph has no gaps. This can be formulated in different ways:

a) Let x0 ∈ I and let ² > 0 . Then there exists a δ > 0 such that for all x ∈ I ∩ (x0 − δ, x0 + δ) we have

|f(x) − f(x0)| < ²

This tells us that the value of f at nearby points is arbitrarily close to the value of f at x0.

13 14 LECTURE 3. VECTOR SPACES OF FUNCTIONS

b) A reformulation of (a) is:

lim f(x) = f(x0) x→x0 3.2 Space of continuously differentiable func- tions C1(I)

Example (2). The space C1(I). Here we assume that I is open. Recall that f is differentiable at x0 if

f(x) − f(x0) f(x0 + h) − f(x0) 0 lim = lim =: f (x0) x→x0 x − x0 h→0 h

0 exists. If f (x0) exists for all x0 ∈ I, then we say that f is differentiable on I. In this case we get a new function on I

x 7→ f 0(x)

We say that f is continuously differentiable on I if f 0 exists and is con- tinuous on I. Recall that if f and g are differentiable, then so are f + g and rf (r ∈ R) moreover (f + g)0 = f 0 + g0 ;(rf)0 = rf 0 As f 0 + g0 and rf 0 are continuous by Example (1), it follows that C1(I) is a vector space.

3.2.1 A continuous but not differentiable function Let f(x) = |x| for x ∈ R. Then f is continuous on R but it is not differentiable on R. We will show that f is not differentiable at x0 = 0. For h > 0 we have f(x + h) − f(x ) |h| − 0 h 0 0 = = = 1 h h h hence f(x + h) − f(x ) lim 0 0 = 1 h→0+ h 3.3. SPACE OF R-TIMES CONTINUOUSLY DIFFERENTIABLE FUNCTIONS CR(I)15

But if h < 0, then

f(x + h) − f(x ) |h| − 0 −h 0 0 = = = −1 h h h hence f(x + h) − f(x ) lim 0 0 = −1 h→0− h

Therefore, f(x + h) − f(x ) lim 0 0 h→0 h does not exist.

3.3 Space of r-times continuously differentiable functions Cr(I)

Example (3). The space Cr(I) Let I = (a, b) be an open interval. and let r ∈ N = {1, 2, 3, ···}.

Definition. The function f : I −→ R is said to be r-times continuously differentiable if all the f 0, f 00, ··· , f (r) exist and f (r) : I −→ R is continuous. We denote by Cr(I) the space of r-times continuously differentiable func- tions on I. Cr(I) is a subspace of C(I).

We have Cr(I) $ Cr−1(I) ( ··· $ C1(I) $ C(I).

3.3.1 Cr(I) 6= Cr−1(I) We have seen that C1(I) 6= C(I). Let us try to find a function that is in C1(I) but not in C2(I).

5 Assume 0 ∈ I and let f(x) = x 3 . Then f is differentiable and

5 2 0 3 f (x) = 3 x 16 LECTURE 3. VECTOR SPACES OF FUNCTIONS which is continuous.

If x 6= 0, then f 0 is differentiable and 10 1 00 − 3 f (x) = 3 x

But for x = 0 we have 0 2 f (h) − 0 5 h 3 5 − 1 lim = lim = lim h 3 h→0 h h→0 3 h h→0 3 which does not exist. Remark. One can show that the function

3r−1 f(x) = x 3 is in Cr−1(R), but not in Cr(R).

Thus, as stated before, we have Cr(I) $ Cr−1(I) ··· $ C1(I) $ C(I).

3.4 Piecewise-continuous functions PC(I)

Example (4). Piecewise-continuous functions on I

Definition. Let I = [a, b). A function f : I −→ R is called piecewise-continuous if there exists finitely many points

a = x0 < x1 < ··· < xn = b such that f is continuous on each of the sub-intervals (xi, xi+1) for i = 0, 1, ··· , n − 1.

Remark. If f and g are both piecewise-continuous, then f + g and rf (r ∈ R) are piecewise-continuous.

Hence the space of piecewise-continuous functions is a vector space. We denote this vector space by PC(I). 3.5. THE χA 17

3.5 The indicator function χA

Important elements of PC(I) are the indicator functions χA, where A ⊆ I is a sub-interval. Let A ⊆ R be a . Define ½ 1, if x ∈ A χ (x) = A 0, if x∈ / A

So the values of χA tell us whether x is in A or not. If x ∈ A, then χA(x) = 1 and if x∈ / A, then χA(x) = 0.

We will work a lot with indicator functions so let us look at some of their properties.

3.5.1 Some properties of χA Lemma. Let A, B ⊆ I. Then

χA∩B(x) = χA(x) χB(x)(∗) Proof. We have to show that the two functions

x 7→ χA∩B(x) and x 7→ χA(x)χB(x) take the same values at every x ∈ I. So lets evaluate both functions: If x ∈ A and x ∈ B, that is x ∈ A ∩ B, then χA∩B(x) = 1 and,

since χA(x) = 1 and χB(x) = 1, we also have χA(x)χB(x) = 1. Thus, the left and the right hand sides of (∗) agree. On the other hand, if x∈ / A ∩ B, then there are two possibilities:

• x∈ / A, then χA(x) = 0, so χA(x)χB(x) = 0.

• x∈ / B, then χB(x) = 0, so χA(x)χB(x) = 0.

It follows that 0 = χA∩B(x) = χA(x)χB(x) 18 LECTURE 3. VECTOR SPACES OF FUNCTIONS

3.5.2 χA∪B with A, B disjoint

What about χA∪B? Can we express it in terms of χA and χB? If A and B are disjoint, that is A ∩ B = ∅ then

χA∪B(x) = χA(x) + χB(x).

Let us prove this:

• If x∈ / A ∪ B, then x∈ / A and x∈ / B. Thus the LHS (left hand side) and the RHS (right hand side) are both zero.

• If x ∈ A ∪ B then either

x is in A but not in B. In this case

χA∪B(x) = 1 and χA(x) + χB(x) = 1 + 0 = 1 or x is in B but not in A. In this case

χA∪B(x) = 1 and χA(x) + χB(x) = 0 + 1 = 1

3.5.3 χA∪B with A, B arbitrary Thus we have,

If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).

Now, what if A ∩ B 6= ∅?

Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x). Proof. • If x∈ / A ∪ B, then both of the LHS and the RHS take the value 0.

• If x ∈ A ∪ B, then we have the following possibilities:

1. If x ∈ A, x∈ / B, then χA∪B(x) = 1 χA(x) + χB(x) − χA∩B = 1 + 0 − 0 = 1 2. Similarly for the case x ∈ B, x∈ / A: LHS equals the RHS. 3.6. 19

3. If x ∈ A ∩ B, then χA∪B(x) = 1 χA(x) + χB(x) − χA∩B = 1 + 1 − 1 = 1

As we have checked all possibilities, we have shown that the statement in the lemma is correct

3.6 Exercise

Write the product

(2χ[0,3] − 3χ[3,6])(χ[0,2] + 5χ[2,4] − 7χ[4,6]) as a linear combination of the indicator functions of intervals.