prove that a function is an inner product

Part (i)

If you ever want to show something is an inner product, you need to show three things for all $f, g \in V$ and $\alpha \in \mathbb{R}$:

1. Symmetry: $\newcommand{\inp}[1]{\left\langle #1 \right\rangle}$ $\inp{f, g} = \inp{g, f}$ (Or, if the field is the complex numbers, $\inp{f,g} = \overline{\inp{g,f}}$, i.e. "conjugate symmetry.)

2. Linearity: $\inp{\alpha f, g} = \alpha \inp{f, g}$. Notice this also implies $\inp{f, \alpha g} = \alpha \inp{f, g}$ ($\overline{\alpha}$ in the complex case) by symmetry.

3. Positive-definite: $\inp{f, f} \ge 0$ with equality if and only if $f = 0$, the zero function.

The first two properties follow directly from the definition of an integral. For the third property, you have $$ \int_a^b f^2 + \int_a^b (f')^2 \ge 0 $$ Now when is this equal to $0$? Well, recall that if a continuous function is positive anywhere, then the integral is positive. Since $f^2$ is continuous, this means $f = 0$ everywhere. Thus $f' = 0 $ everywhere also, so equality holds.

Part (ii)

If you ever want to show something is a norm, you need to show three things for all $f, g \in V$ and $\alpha \in \mathbb{R}$:

1. Scales in absolute value: $\newcommand{\norm}[1]{\left\| #1 \right\|}$ $\norm{\alpha f} = |\alpha| \norm{f}$.

2. Triangle Inequality: $\norm{f + g} \le \norm{f} + \norm{g}$.

3. Separates Points: $\norm{f} = 0$ if and only if $f = 0$.

Again the first two properties follow directly from the definition of the norm, which is an integral. For the third property, we use the same property of integrals we used before: if a continuous function is positive anywhere, its integral is positive. Since $|f|$ is continuous when $f$ is, this means $\int |f| = 0$ if and only if $f = 0$. Which in turn implies $\int |f'| = 0$.

You will notice parts (i) and (ii) seem very similar. It's almost the case that you can use part (i) for part (ii), but not quite. The problem is that $\inp{f,f}$ is not equal to $\norm{f}^2$ as expected.