Which of the following are subspaces of the vector space of real-valued functions of a real...

Which of the following are subspaces of the vector space of real-valued functions of a real variables? (must select all of the subspaces.)

A. The set of even function (f(-x) = f(x) for all numbers x).

B. The set of odd functions (f(-x) = -f(x) for all real numbers x).

C. The set of functions f such that f(0) = 7

D. The set of functions f such that f(7) = 0

Solutions

Expert Solution

A. Let S be the set of all even functions. Now, if and g are even functions, then (f+g)(-x) = f(-x)+g(-x) = f(x)+g(x) = (f+g)(x) so that f+g is an even function i.e. S is closed under vector addition. Further, if k is a scalar, then kf(-x) = k(f(-x))= kf(x) so that kf is an even function i.e. S is closed under scalar multiplication. Also, the zero function is apparently in S so that S is a vector space, and hence a subspace of the space of all real valued functions of real variables.

B. Let S be the set of all odd functions. Now, if and g are odd functions, then (f+g)(-x) = f(-x)+g(-x) = -f(x)-g(x) = -(f+g)(x) so that f+g is an odd function i.e. S is closed under vector addition. Further, if k is a scalar, then kf(-x) = k(f(-x))= k(-f(x)) = -kf(x) so that kf is an odd function i.e. S is closed under scalar multiplication. Also, the zero function is apparently in S so that S is a vector space, and hence a subspace of the space of all real valued functions of real variables.

C. Let f and g be 2 arbitrary functions in the given set S (say). Then f+g(0) = f(0)+g(0) = 7+7= 14 so that f+g is not in S i.e. S is not closed under vector addition Hence S is not a vector space, and hence not a subspace of the space of all real valued functions of real variables.

D. Let f and g be 2 arbitrary functions in the given set S (say). Then f+g(7) = f(7)+g(7) = 0+0= 0 so that f+g is in S i.e. S is closed under vector addition. Further, if k is a scalar, then kf(7)= k(f(7)) = k*0= 0 so that kf is in S i.e. S is closed under scalar multiplication. Also, the zero function is apparently in S so that S is a vector space, and hence a subspace of the space of all real valued functions of real variables.

milcah answered 1 year ago

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