I was wondering what the solution to number 4 on the last exam is? I got partial credit, but I'm still feeling uneasy with my answer.
$$ \langle f,g \rangle = \int_0^1 f(x)g(x)\,dx$$
Given any three such functions $f,g,$ and $h$, prove that
$$ \langle f - g, h \rangle = \langle f,h \rangle - \langle g,h \rangle. $$
@jtamberi's answer here should help. The proof there is VERY similar proof for this problem if you let $a=1$ and $b=-1$. Just keep in mind you are using different inner products.
$$\langle f-g,h\rangle$$
$$\int_{0}^{1}(f(x)-g(x))h(x)dx$$
$$\int_{0}^{1}f(x)h(x)-g(x)(hx)dx$$
$$\int_{0}^{1}f(x)h(x)dx-\int_{0}^{1}g(x)(hx)dx$$
$$\langle f,h\rangle-\langle g,h\rangle$$
@jradford Shouldn't the final line be:
$$\langle f,h \rangle - \langle g,h \rangle\ $$
that you've proved equal to $\langle f-g,h \rangle $?
You're right, my bad! I edited my post.
How many points was this question worth? I only got 8, so if it is out of 10 I don't understand what I am missing. I have the same thing that @jradford has.
Mark did put a "? =" on the outside of this integral.
$$\int_{0}^{1}(f(x)-g(x))h(x)dx$$
Was I supposed to set it equal to something? I didn't start off with < f-g,h>, I kind of just dove right into the proof. Could that have be my error?
I would guess that that not starting with $\langle f - g,h\rangle =$ could have been the problem. The idea with a proof is to show explicitly through simple steps that A implies (or equals) B. The problem was to show that $$\langle f - g, h \rangle = \langle f,h \rangle - \langle g,h \rangle$$ If you had everything the same as the proof above except for the first line, then what you showed was $$\int_{0}^{1}(f(x)-g(x))h(x)dx=\langle f,h \rangle - \langle g,h \rangle$$ You and I know that $\langle f - g,h\rangle=\int_{0}^{1}(f(x)-g(x))h(x)dx$ but since you did not include that you did not actually prove the equality of the two inner products.
Thank you so much! I actually had that and then freaked out on the test for some reason...
Proofs are typically just a matter of being excessively verbose... well if only it was always that easy.