Be the Real Vector Space of Continuous Functions Defined on
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The real-valued continuous functions on the real numbers form a vector space C(R) {f : R 5 R| f is continuous} with: Addition (f + 9)(x) = f(z) + g(x). Scalar multiplication (kf)(x) = kf(z) Is the set S = {f e C(R) |f(3) = 0} subspace of the vector space C(R)? Is the set S = {f € C(R) |f(3) = 2} subspace of the vector space C(R)?
Related Question
Let C[a b] be the set of all continuous real valued functions On the interval [a . 6]. If f is in Cla,b]. we can extend continuous function from R to R by letting be the function defined by f (a) if I < ( F(c) f(z) if a <r <b. (f() if b < € this way we can view C[a,b] as subset of F the vector space of all functions from R to Verify that C[a,b] is a vector space_
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Video Transcript
problem. We are asked to show that the space of continuous functions on the interval A to B is a vector space when we view it as a subset of capital F, which is the vector space of functions from the real numbers to the real numbers. And were given a particular way of extending continuous functions on the interval from A to B. Two functions from the reels to the reels. The first thing we need to check is that the zero factor is included in this subset of F. So can we take the zero function on a B and extended to the zero function F. Well, yes, we can we just take F of X is equal to zero if X is less than a zero if um A is less than or equal to X is less than or equal to B and zero if B is less than X. And this is an extension of the zero function on the interval A B to the zero function on the real numbers. And so C A B contains the zero function. The next thing we need to show is that this set is closed under linear combinations. So, first of all, if F and G belong to the set of continuous functions on the interval A. B. And let's say R and s are real numbers, then first of all our F plus S G is continuous. So this is an element of the uh set of function, continuous functions on the interval A B. And we could extend it to a function on the, from the reels to the reels simply by saying, well let's call this H of X. Lower case H of X rather. And we could extend this to a function um from the reels to the reels by saying Capital H of X is equal to um A if X is less than a, it's equal to H of X. If A is less than or equal to X and less than or equal to be and it's equal to be if B is less than X. Using the extension method that were given in the problem. And so if F and G belongs to the set of continuous functions, then a linear combination of them can be extended to uh function from the reels to the reels, which shows that C a P is closed under linear combinations. And so this shows that C a P is a subspace of the vector space F. And in particular is a vector space.
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