Verify that this is an inner product. 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 $(f,f)=0\iff f=0$.
 A: Your reasoning is correct, but it can be simplified. Define, for continuous functions $f$ and $g$ on $[0,1]$ (with complex values),
$$
(f,g)_0=\int_0^1 f(t)\overline{g(t)}\,dt
$$
Then you clearly have, for $f,g\in C^1([0,1])$,
$$
(f,g)=(f,g)_0+(f',g')_0
$$
Both terms on the right hand side are well defined, because by assumption $f'$ and $g'$ are continuous. Now it is easy to prove that
$$
\bar{f}^{\,\prime}=\overline{f'}
$$
(the derivative of the conjugate is the conjugate of the derivative). So you just need to prove that
$$
(g,f)_0=\overline{(f,g)}
$$
for $f$ and $g$ continuous.
What about the last part? First prove that $(f,f)_0=0$ implies $f=0$ for $f$ continuous. Then
$$
(f,f)=(f,f)_0+(f',f')_0
$$
and, as both terms in the right hand side are $\ge0$, the condition $(f,f)=0$ implies $(f,f)_0=0$.
A: Perhaps you should say both integrals $\displaystyle\int_{0}^{1}f(t)\overline{g(t)}dt,\displaystyle\int_{0}^{1}f'(t)\overline{g'(t)}dt$ exist, after that the integral defined in $(f,g)$ exists.
