Originally Posted by

**alice8675309** Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

V=P_1(R) with <f,g>$\displaystyle \int_{-1}^{1}$f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

Solution:

Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match $\displaystyle \int_{-1}^{1}$(T*f)(t)(b_0+b_1t)dt=$\displaystyle \int_{-1}^{1}$(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.