But of course, that just says this is NOT a vector space- but you could have stopped after showing it is not closed under addition.

(b) Is a vector space.

Closed under vector addition.

Closed under scalar multiplication.

You need to

**show** these, not just say they are true. If 2x(0)= x(1), 2y(0)= y(1) and z(t)= x(t)+ y(t), then 2z(0)= 2(x(0)+ y(0))= 2x(0)+ 2y(0)= x(1)+ y(1)= z(1) and, for any scalar, a, 2(az(t))= 2(ax(t)+ ay(t))= a(2x(t))+ a(2y(t))= ax(1)+ ay(1)= a(x(1)+ y(1))= az(1).

There exists an element (origin)

, such that

Good! You showed that the 0 function

**is** in this set.

(c) and (d), could need a push in the right direction here.

Suppose x(t) is such that

for [/tex]0\le t\le 1[/tex]. What about (-1)x?

(d) is pretty much like the others. If x(t)= x(1-t), y(t)= y(1-t), and z= x+ y, then z(t)= x(t)+ y(t)= x(1-t)+ y(1-t)= z(1- t), etc.