The question was to classify each statement as true or false with justification:
Define $\langle f, g\rangle$ = $\int_0^1 f'(x)g(x) + f(x)g'(x) \,dx $. Then, $\langle , \rangle$ is an inner product on $P_3(\mathbb{R})$.
My solution below concluded that it was true, however my classmate said the statement was false, and although I didn't completely understand their proof, I am beginning to doubt mine. Can anyone clarify if my justification is legitimate?
Rewrite $\int_0^1 f'(x)g(x) + f(x)g'(x) \,dx $ as $$\int_0^1 f'(x)g(x)\,dx + \int_0^1 f(x)g'(x) \,dx $$
For $\int_0^1 f'(x)g(x)\,dx$, integrate by parts to get $f(x)g(x) - \int_0^1 f(x)g'(x)\,dx$. Substitute this result into the rewritten form of $\langle f, g \rangle$, we get $$<f,g>=f(x)g(x) - \int_0^1 f(x)g'(x)\,dx + \int_0^1 f(x)g'(x) \,dx =f(x)g(x)$$
Then, (proving the three properties of inner products) $\langle f,f\rangle =[f(x)]^2\ge0$, also $\langle f,f\rangle=0$ iff $f(x)=0$ for $x\in[0,1]$. Thus, it is positive definite.
Next, $\langle f,g\rangle=f(x)g(x)=g(x)f(x)=\langle g,f\rangle$ so $\langle f,g\rangle$ is symmetric.
Finally, $$\langle sf+tg,h \rangle=(sf(x)+tg(x))h(x)=s(f(x)h(x))+t(g(x)h(x))=s\langle f,h\rangle+t\langle g,h\rangle $$ So $\langle f,g \rangle$ is bilinear, and is thus an inner product on $P_3R$.
Likely made a silly mistake somewhere or my thinking is wrong, any tips would be helpful!
Your calculation of the inner product is not quite correct; you should have
\begin{align*} \langle f, g \rangle &= \int_0^1 \frac{d}{dx} f(x) g(x) \, dx \\ &= f(x) g(x) \big|_0^1 \\ &= f(1) g(1) - f(0) g(0) \end{align*}
This does not define an inner product, although it is bilinear. For example, any function $f$ that vanishes at both $0$ and $1$ will have zero "norm" - to be explicit, if $f(x) = x(x - 1)$ then $\langle f, f \rangle = 0$.