Let $C[-1,2]$ be the space of all continuous functions $f:[-1,2]\to \mathbb{C}$.
Which of the following define an inner product on $C[-1,2]$ and which do not?
$$\langle f,g\rangle =\int_{-1}^{2}|f(t)+g(t)|dt$$
a. $$\langle f,f\rangle =\int_{-1}^{2}|f(t)+f(t)|dt=2\int_{-1}^{2}|f(t)|dt$$ b. $$\overline{\langle g,f\rangle }= \overline{\int_{-1}^{2}|g(t)+f(t)|dt} = \int_{-1}^{2}\overline{|g(t)+f(t)|}dt = \int_{-1}^{2}\overline{|f(t)+g(t)|}dt=\langle f,g\rangle $$ c. $$\langle \alpha f+\beta g,h\rangle=\int_{-1}^{2}|\alpha f(t) +\beta g(t)+h(t)|dt\neq\alpha \int_{-1}^{2}|f(t)+h(t)|dt+\beta \int_{-1}^{2}|g(t)+h(t)|dt=\alpha\langle f,h\rangle+\beta\langle g,h\rangle$$
Can we be sure that the expression in a. is $\geq 0$? as the function can be decreasing? so the overall expression will be less than $0$
Can we move the conjugate inside the integral as in b.?
No, it is not an inner product. For instance, take $f(x)=g(x)=x$ and $\alpha=-1$. Then$$\langle\alpha f,g\rangle=\int_{-1}^2|-x+x|\,\mathrm dx=0$$ and$$\alpha\langle f,g\rangle=-\int_{-1}^2|2x|\,\mathrm dx\neq0.$$So, in general it is not true that $\langle\alpha f,g\rangle=\alpha\langle f,g\rangle$.