On the space $C^1([0,1])$ I want to show that $(f,g)=\int_0^1(f\overline g+f'\overline {g'})$, where the overline denotes the complex conjugate, defines an inner product. Below is my working for the parts I am most unsure of, and I would like to check if what I have done is correct.
First we establish that $(f,g)$ is well defined. For $t\in[0,1]$ consider,
$$(f,g)=\int_0^1(f\bar g+f'\overline {g'})(t)dt$$
$$=\int_0^1f(t)\overline{g(t)}dt+\int_0^1f'(t)\overline{g'(t)}dt$$
Since $f,g\in C^1([0,1])$, and by the fact that the product of continuous functions is itself continuous, then both integrations are meaningful, and thus, $(f,g)\in\mathbb R$.
Secondly, we establish conjugate symmetry. This is the criterion of an inner product that I am, perhaps, most unsure of. We have that,
$$(f,g)=\int_0^1f(t)\overline{g(t)}dt+\int_0^1f'(t)\overline{g'(t)}dt$$
We use the property that for any $a\in\mathbb R,\,\overline a=a$. For all $t\in[0,1]$, the products $f(t)\overline {g(t)}$ and $f'(t)\overline{g'(t)}$ define real numbers. As such, $\overline {f(t)\overline {g(t)}}=\overline {f(t)}g(t)$ and $\overline{f'(t)\overline{g'(t)}}=\overline{f'(t)}g'(t)$. Thus, we have that,
$$=\int_0^1\overline{f(t)}{g(t)}dt+\int_0^1\overline{f'(t)}g'(t)dt$$
And by the commutativity of multiplication of real numbers,
$$=\int_0^1g(t) \overline{f(t)}dt+\int_0^1g'(t)\overline{f'(t)}dt$$
$$=\overline{(g,f)}$$
Is this correct, or is there anything I could improve upon? I'm fairly confident in proving that $(f,f)\ge0$ and that $(f,f)=0\iff f=0$.
