Question

.B Consider the vector space V of all function from R to R. Define T:V→V by setting T(f)(t)= 1/2 f(t)+ 1/2f(−t) for every f∈V and every t∈R. (a) Show that T is a projection. (b) Show that Range (T) is the set of all even functions in V, and that Null(T) is the set of all odd functions in V. (See below for definitions.)

Answer 1

T is a projection because T(T(f)) = T(f) for every f in V. The range of T is the set of all even functions, as shown by the property f(t) = f(-t) for functions in the range. The null space of T is the set of all odd functions, characterized by the property f(t) = -f(-t).

Step-by-step explanation:

To demonstrate that T is a projection in the vector space V of all functions from ℝ to ℝ, we need to show that T(f) satisfies the condition T(T(f)) = T(f) for every function f in V. Applying T twice, we get T(T(f))(t) = ½(½ f(t) + ½ f(-t)) + ½(½ f(-t) + ½ f(t)) = ½ f(t) + ½ f(-t) = T(f)(t), thus confirming that T is indeed a projection.

The range of T can be shown to be the set of all even functions. If f is in the range of T, then there exists some function g in V such that T(g) = f. Therefore, f(t) = T(g)(t) = ½ g(t) + ½ g(-t), and we can see that f(t) = f(-t), satisfying the property of even functions. Conversely, for any even function f, T(f) = f. This shows that the range of T consists precisely of even functions.

The null space of T, or Null(T), is the set of functions f such that T(f) = 0. For a function f to be in Null(T), the condition ½ f(t) + ½ f(-t) must equal zero, which happens if and only if f(t) = -f(-t); that is, f is an odd function. Thus, Null(T) is the set of all odd functions in V.