Let C^infinity denote the vector space of all functions f : R-->R whose derivatives f^n exist for all n>=0.

Prove or disprove that it is a linear transformation


a) T : R^2 --%26gt;R^2 with T(x,y) = (x - y, 1 - y + x)





b) T : C^infinity --%26gt; C^ infinity with T(f) = f" - 3f' + 2f.





c) T : C^infinity --%26gt; R where T(f) = f(2).





d) T : C^infinity --%26gt; C^infinity where T(f) = the integral form 0 to x of f(t) dt + 1.





e) T : C^infinity --%26gt; C^infinity with T(f) = f(x + 1)

Let C^infinity denote the vector space of all functions f : R--%26gt;R whose derivatives f^n exist for all n%26gt;=0.
I'm not too sure about these, but here goes:





If any of these are linear transformations, we must have


T(f+g)=T(f)+T(g)


T(af)=a*T(f)


for all functions f and g in C^inf and scalars a.





a) Well, this one isn't about our functions yet. The same basic thing applies though:


T( (x,y) + (u,v) ) = T( (x+u,y+v) ) = (x+u-y-v, 1-y-v+x+u)


= (x-y, 1-y+x) + (u-v, -v+u).


But we cannot get another 1 into the second ordered pair to obtain T((x,y))+T((u,v)), so this is not linear.





b) This one is linear, following from the linear properties of the derivative:


T(f+g) = (f+g)'' - 3(f+g)' + 2(f+g)


= f'' + g'' - 3f' - 3g' + 2f + 2g = f''-3f'+2f + g''-3g'+2g = T(f)+T(g)


T(af) = (af)'' - 3(af)' + 2(af) = af'' - 3af' + 2af = a*T(f)





c) This is linear as well, and follows rather simply from the definition of sums and scalar products of functions. I'll leave the details to you.





d) This fails because of the +1:


T(f+g) = int( (f+g)(t) ) + 1 = int( f(t)+g(t) ) +1


= int(f(t)) + int(g(t)) + 1


and we have our problem.





e) Linear again:


T(f+g) = (f+g)(x+1) = f(x+1)+g(x+1) = T(f)+T(g)


T(af) = (af)(x+1) = a*f(x+1) = a*T(f).





If you notice anything wrong in what I've done (this goes for other answerers as well obviously), please let me know.





Cheers!