Math 328 Course Notes Ian Robertson March 3, 2006 3 Properties of C[0, 1]: Sup-norm and Completeness In this chapter we are going to examine the vector space of all continuous functions defined on the closed interval [0,1]. The notions of sup-norms, completeness, and different types of convergence will be seen. 3.1 Pointwise vs. Uniform Convergence Definition: Suppose that I is an interval in R and that f 1, f 2,... are real-valued functions defined on I. For each point x I we have a sequence (f n (x)) n 1 of numbers. When we fix a value of x I we can ask whether the limit lim n f n (x) exists. If the limit exists for each x, taken individually, then we say the sequence converges pointwise. Definition: Let (f n ) n 1 be a sequence of functions defined from [0, 1] to R. This sequence converges uniformly to some function f if for any ɛ > 0 there is an N with the property that: f n (x) f(x) < ɛ n N, x [0, 1] You can see that for a sequence to converge uniformly, the N picked must be the same for all values of x, whereas pointwise convergence only requires that a value for N is able to be found for each fixed x. Clearly uniform convergence implies pointwise convergence, but the converse is not true in general. 3.2 Theorem - Uniform Limit of Continuous Functions Let (f n ) n 1 be a sequence of functions defined from [0, 1] to R, converging uniformly to f. If all functions f n are continuous, then f is also continuous. (Note, this is not true if we only have pointwise convergence.) Proof: Let ɛ > 0 be given. We need to show that δ > 0 such that f(x) f(y) < ɛ whenever x y < δ. (x, y [0, 1]) We know by the triangle inequality on R that: f(x) f(y) f(x) f n (x) + f n (x) f n (y) + f n (y) f(y) 1
Because of uniform convergence, we can pick an N such that f n (x) f(x) < ɛ 3 n N, x [0, 1]. Now consider f N. By continuity, we know that we can pick a δ such that: f N (x) f N (y) < ɛ 3 whenever x y < δ, Combining these two facts together, we know that δ > 0 such that: f(x) f(y) < ɛ whenever x y < δ But what about a sequence of functions that converges pointwise to some function? Let s look at an example of a pointwise-convergent sequence: f n (x) = x n Clearly, if we fix x in the interval [0, 1], the sequence (f n (x)) n 1 converges to 0 if x 1 and to 1 if x = 1. So here we have a sequence of continuous functions (polynomials are always continuous) which converges pointwise to a piece-wise function, which is not continuous at x = 1. 3.3 The Sup-norm Definition: For f C[0, 1], we define the sup-norm of f as follows: f sup-norm(f) sup f(x) Recall: the supremum of a set is the least upper bound of the set, a number M such that no element of the set exceeds M, but for any positive ɛ, there is a member of the set which exceeds M ɛ. 3.4 C[0, 1] is a vector space It is possible to think of C[0, 1] as a vector space over the real numbers, with an addition and multiplication by scalars defined as such: (f + g) : x f(x) + g(x) (α f) : x α f(x) for α R. Most of the vector space properties are inherited from the fact that R is a vector space as well. We set f(x) = 0, i.e. x [0, 1], f 0 as the additive identity for this vector space, and α = 1 as the scalar multiplicative identity. It is clear that all of the other vector space properties hold for C[0, 1]. We can now think of functions f in C[0, 1] as vectors in an abstract vector space, which is a powerful notion. 2
3.5 C[0, 1] is a normed vector space Proposition:, the sup-norm, is a norm on C[0, 1]. For this to be true, we need the following to hold: 1. f 0, f [0, 1] 2. f = 0 f = 0 3. λ f = λ f 4. f + g f + g The proof of this list is fairly straightforward: 1. This property is clear, using the definition of the sup-norm. The sup-norm is the largest value of a set of absolute values, so it is obvious that it must be greater than or equal to zero. 2. The direction is simple to see, if f(x) 0, sup f(x) is clearly equal to zero. The direction is also quite easy to see. If sup f(x) = 0, then f(x) = 0 for every x in [0, 1]. 3. λ f = sup λ f(x). By properties of the supremum, we can bring the λ outside of the absolute brackets, and get: λ f = λ sup f(x) = λ f 4. f + g = sup f(x) + g(x). We know by the triangle inequality on R that: f(x) + g(x) f(x) + g(x) and so therefore, sup f(x) + g(x) sup f(x) + sup g(x) So now we can use the fact that C[0, 1] is a normed vector space. 3.6 Convergence and the Sup-norm Theorem: Let f and f n, n N, be continuous functions in C[0, 1]. Then: (f n ) n 1 converges uniformly to f lim n f n f = 0 Proof: The direction is pretty easy to see. Because of the uniform convergence of (f n (x)) n 1, we know that for any ɛ greater than zero, we can find an N such that for all x in the interval, f n (x) f(x) < ɛ, n N. 3
Using this fact, and since it s easy to see that: f n (x) f(x) < ɛ x sup f n (x) f(x) < ɛ ɛ > 0, N such that: sup f n (x) f(x) < ɛ, n N. The direction is an easy reversal of argument, so it is left as an exercise. 3.7 Some Definitions Definitions: For f n C[0, 1], n N: 1. (f n ) n 1 converges to f C[0, 1] if: ɛ > 0, N N such that f n f < ɛ, n N 2. (f n ) n 1 is a Cauchy sequence if: ɛ > 0, N N such that f n f m < ɛ, n, m N 3.8 Results of the Triangle Inequality Remark: It is obvious (by the Triangle Inequality) that: 1. There can be at most one limit of a convergent sequence of functions in C[0, 1] 2. Every convergent sequence in C[0, 1] is Cauchy. Proof: 1. Assume that (f n ) converges to f, and also to g. What is f g? By the triangle inequality: for large n {}}{ f g = f f n + f n g f f n }{{} + g f n ] for large n f g < ɛ, ɛ > 0 f g = 0 f g = 0 f = g 4
2. Assume that (f n ) converges to f. What can we say about f m f n? By the triangle inequality, for any ɛ > 0: f m f n = f m f + f f n f m f n < ɛ, for big enough m, n 3.9 The Completeness of C[0, 1] for large n {}}{ f m f }{{} + f f n ] for large m It should be noted that vector spaces on which we want to carry out any kind of meaningful analysis must be complete. We have already seen the completeness of R, and all of the analysis that can be completed on the real numbers. The logical question to ask is whether the vector space of continuous functions is complete. Theorem: C[0, 1] is complete. To have completeness of a vector space, we need to show that every Cauchy sequence in that space converges to an element of the space. Proof: Consider a Cauchy sequence (f n ) n 1 in C[0, 1]. This means that: ɛ > 0, N N such that: f m f n < ɛ, n, m N We need to show that for such an f, ɛ > 0, N N such that: f n (x) f(x) < ɛ, n N. For each fixed x in the interval, we have that f n (x) f m (x) < ɛ, for m, n greater than some N. Thus each (f n (x)) n N is a Cauchy sequence in R. Now since R is complete, there is a limit of this sequence in R. We ll call this f(x) = lim x f n (x). Claim: f n converges to f uniformly, because: f(x) f m (x) = lim n f n(x) f m (x) Now since f n is a Cauchy sequence, there is a value of N, independent of x, such that f n (x) f m (x) < ɛ, n, m N f(x) f m (x) ɛ, m N, x [0, 1] f m f uniformly 5
Since f is the uniform limit of continuous functions, f itself is continuous (from 3.2), so: (f n ) n 1 converges in C[0, 1] to f. So now we have seen that C[0, 1] is a complete, normed vector space. We can now think of two functions f and g as vectors in an abstract vector space, with a notion of distance between the two functions given by the sup-norm of the difference f g. In the next chapter we ll investigate the compactness of C[0, 1], and the idea of equicontinuity. 6