This problem is from Elementary Linear Algebra 9th Edition by Howard Anton. Prob

Question

This problem is from Elementary Linear Algebra 9th Edition by Howard Anton. Problem 4)a

Use theorem 5.2.1 to determine which of the following are subspaces of the space F(-infinity,infinity):

Problem: All f such that f(x)<=0 for all x. Here is the theorem: a) If u and v are vectors in W, then u+v is in W. b) If k is any scalar and u is any vector in W, then ku is in W.
Now, my first question: F(-infinity,infinity) [with a capital f] denotes all functions from x=-infinity to infinity , correct? Now to check part a of the theorem,
f(x)<=0 g(x)<=0 f(x)+g(x)<=0 Therefore, this set is closed under addition. To check part b, k*f(x)<=0 However, if k is negative, f(x) will be >=0. Therefore, this set isn't closed under scalar multiplication. Would you say this answer is correct? The solution here on Cramster sounds wrong to me.
Problem: All f such that f(x)<=0 for all x. Here is the theorem: a) If u and v are vectors in W, then u+v is in W. b) If k is any scalar and u is any vector in W, then ku is in W.
Now, my first question: F(-infinity,infinity) [with a capital f] denotes all functions from x=-infinity to infinity , correct? Now to check part a of the theorem,
f(x)<=0 g(x)<=0 f(x)+g(x)<=0 Therefore, this set is closed under addition. To check part b, k*f(x)<=0 However, if k is negative, f(x) will be >=0. Therefore, this set isn't closed under scalar multiplication. Would you say this answer is correct? The solution here on Cramster sounds wrong to me.

Explanation / Answer

sorry! i have seen the question but not completely gone through the notes you wrote. but, the question itself is self contained. a vector space is an additive group. you gave that the set is containing all the functions which take only < = 0 images. now, suppose A is the subset of the said functions. i.e. A contains functions which take images < = 0 if f(x) < = 0, then there must be a function g in A which is the additive inverse of f. that means (f+g)(x) = 0(x) = 0 ==> g(x) must be greater than or equal to zero. in other words, the additive inverse of f is g provided g takes images greater than or equal to zero. but such g's are not available in A. therefore , A is not a subspace of the space of continuous functions.

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