12 Lecture 3 Vector Spaces of Functions 3.1 Space of Continuous Functions C(I) Let I ⊆ R be an interval. Then I is of the form (for some a < b) 8 > fx 2 Rj a < x < bg; an open interval; <> fx 2 Rj a · x · bg; a closed interval; I = > fx 2 Rj a · x < bg :> fx 2 Rj a < x · bg: Recall that the space of all functions f : I ¡! R is a vector space. 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 2 R). Hence C(I) is a vector space. Recall, that a function is continuous, if the graph has no gaps. This can be formulated in di®erent ways: a) Let x0 2 I and let ² > 0 . Then there exists a ± > 0 such that for all x 2 I \ (x0 ¡ ±; x0 + ±) we have jf(x) ¡ f(x0)j < 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 di®erentiable func- tions C1(I) Example (2). The space C1(I). Here we assume that I is open. Recall that f is di®erentiable 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 2 I, then we say that f is di®erentiable on I. In this case we get a new function on I x 7! f 0(x) We say that f is continuously di®erentiable on I if f 0 exists and is con- tinuous on I. Recall that if f and g are di®erentiable, then so are f + g and rf (r 2 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 di®erentiable function Let f(x) = jxj for x 2 R. Then f is continuous on R but it is not di®erentiable on R. We will show that f is not di®erentiable at x0 = 0. For h > 0 we have f(x + h) ¡ f(x ) jhj ¡ 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 ) jhj ¡ 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 di®erentiable functions Cr(I) Example (3). The space Cr(I) Let I = (a; b) be an open interval. and let r 2 N = f1; 2; 3; ¢ ¢ ¢ g. De¯nition. The function f : I ¡! R is said to be r-times continuously di®erentiable if all the derivatives 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 di®erentiable 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 ¯nd a function that is in C1(I) but not in C2(I). 5 Assume 0 2 I and let f(x) = x 3 . Then f is di®erentiable 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 di®erentiable 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 De¯nition. Let I = [a; b). A function f : I ¡! R is called piecewise-continuous if there exists ¯nitely 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 2 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 INDICATOR FUNCTION ÂA 17 3.5 The indicator function ÂA Important elements of PC(I) are the indicator functions ÂA, where A ⊆ I is a sub-interval. Let A ⊆ R be a set. De¯ne ½ 1; if x 2 A Â (x) = A 0; if x2 = A So the values of ÂA tell us whether x is in A or not. If x 2 A, then ÂA(x) = 1 and if x2 = A, then ÂA(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 ÂA Lemma. Let A; B ⊆ I. Then ÂA\B(x) = ÂA(x) ÂB(x)(¤) Proof. We have to show that the two functions x 7! ÂA\B(x) and x 7! ÂA(x)ÂB(x) take the same values at every point x 2 I. So lets evaluate both functions: If x 2 A and x 2 B, that is x 2 A \ B, then ÂA\B(x) = 1 and, since ÂA(x) = 1 and ÂB(x) = 1, we also have ÂA(x)ÂB(x) = 1. Thus, the left and the right hand sides of (¤) agree. On the other hand, if x2 = A \ B, then there are two possibilities: ² x2 = A, then ÂA(x) = 0, so ÂA(x)ÂB(x) = 0. ² x2 = B, then ÂB(x) = 0, so ÂA(x)ÂB(x) = 0. It follows that 0 = ÂA\B(x) = ÂA(x)ÂB(x) 18 LECTURE 3. VECTOR SPACES OF FUNCTIONS 3.5.2 ÂA[B with A, B disjoint What about ÂA[B? Can we express it in terms of ÂA and ÂB? If A and B are disjoint, that is A \ B = ; then ÂA[B(x) = ÂA(x) + ÂB(x): Let us prove this: ² If x2 = A [ B, then x2 = A and x2 = B. Thus the LHS (left hand side) and the RHS (right hand side) are both zero. ² If x 2 A [ B then either x is in A but not in B. In this case ÂA[B(x) = 1 and ÂA(x) + ÂB(x) = 1 + 0 = 1 or x is in B but not in A. In this case ÂA[B(x) = 1 and ÂA(x) + ÂB(x) = 0 + 1 = 1 3.5.3 ÂA[B with A, B arbitrary Thus we have, If A \ B = ;, then ÂA[B(x) = ÂA(x) + ÂB(x): Now, what if A \ B 6= ;? Lemma. ÂA[B(x) = ÂA(x) + ÂB(x) ¡ ÂA\B(x): Proof. ² If x2 = A [ B, then both of the LHS and the RHS take the value 0. ² If x 2 A [ B, then we have the following possibilities: 1. If x 2 A; x2 = B, then ÂA[B(x) = 1 ÂA(x) + ÂB(x) ¡ ÂA\B = 1 + 0 ¡ 0 = 1 2. Similarly for the case x 2 B; x2 = A: LHS equals the RHS. 3.6. EXERCISE 19 3. If x 2 A \ B, then ÂA[B(x) = 1 ÂA(x) + ÂB(x) ¡ ÂA\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 (2Â[0;3] ¡ 3Â[3;6])(Â[0;2] + 5Â[2;4] ¡ 7Â[4;6]) as a linear combination of the indicator functions of intervals..