Final answer:
The set of continuous functions where f(0)=−1 is not a vector space because it does not contain the null vector, and it fails the closure property under vector addition and scalar multiplication. Therefore, it is not a subspace of V, the space of all continuous functions on [−1,1].
Step-by-step explanation:
To determine whether a set of continuous functions on the interval [−1,1] where f(0)=−1 is a vector space, we need to see whether this set satisfies the vector space axioms, such as closure under addition and scalar multiplication, and includes the null vector. The null vector in this context would be the zero function, which is a function that is equal to zero at every point in its domain. However, since f(0) = −1, we do not have the zero function in this set, because the zero function would require f(0) to be 0, not −1. Consequently, this set does not contain the null vector and therefore is not a vector space. Additionally, since the null vector is absent, the set cannot be a subspace of the vector space V of all continuous functions on [−1,1].
Moreover, if we consider any two functions f and g from this set and any scalar c, the function (cf + g) must also satisfy the condition (cf + g)(0) = −1. However, this may not be true for all choices of c, f, and g, which means this set fails the closure property and thus definitely cannot be a subspace of V.