[Apostol's Calculus] Exercises 1.5: Solution to 5

Let S be the set of all functions f such that f(1) = 1 + f(0). Assume that S is a real linear space. Then S must be closed under addition. In other words, if g(x) is in S, then the following must hold:

g(x) ∈ S
(g(x) + g(x)) ∈ S
2g(x) ∈ S

Now let's define the following:

(2g(x) = h(x))

Notice that since we're assuming that S is a real linear space, then h(x) must be in S, since real linear spaces are closed under addition. Furthermore, we were told that all functions f in S must satisfy the property f(1) = 1 + f(0). Thus,

h(1) = 1 + h(0)

But h(x) = 2g(x). So,

2g(1) = 1 + 2g(0)
g(1) = 0.5 + g(0)

But this violates the condition of being in S, and we assumed that g(x) was in S. So S cannot possibly be closed under addition.

It follows that S cannot possibly be a real linear space, since it is not closed under addition.

QED