Interpretation of zero angle between two elements in a inner product space Take $f,g \in V$, where $V$ is an inner product space. Let $\langle \cdot, \cdot \rangle : V \times V \to [0,\infty)$ denote the inner product operator in $V$. Let the "angle" $\theta$ between $f$ and $g$ be defined through the rule
$$
\cos(\theta) = \frac{\langle f, g\rangle}{{\left\|f\right\| \left\|g\right\|}}
$$
where norms on $V$ are defined in terms of the inner product as $\left\|\cdot\right\| \doteq \langle \cdot, \cdot \rangle $.
My question is simple: if $\cos(\theta) = 1$, what conclusions can be made? In particular, I would like to know if I can conclude that $f = g$ almost everywhere, and if not, I would like to know what extra assumptions are needed to get that result. In particular, I am interested to find out if $f = g \text{ }\mathrm{a.e.}$ when $\cos(\theta) = 1$ for the restricted case when $V$ is the space of bounded, real valued functions whose domain is a closed interval in the real line.
Thank you very much for your help!
 A: (The question is tagged with [real-analysis], therefore I'll assume that $V$ is an inner product space over $\Bbb R$.)
If
$$ \tag{*}
\frac{\langle f, g\rangle}{{\|f\| \|g\|}} = 1
$$
then $f$ and $g$ are necessarily non-zero, and $\langle f, g\rangle$ is a positive real number. It follows that
$$
|\langle f, g\rangle| = \langle f, g\rangle = \|f\| \|g\| \, ,
$$
i.e. we have equality in the Cauchy–Schwarz inequality, which is the case if and only if $f$ and $g$ are linearly dependent. Since both are non-zero, we have
$$
 f = c g \text{ for some } c \in \Bbb R
$$
and since $\langle f, g\rangle  > 0$
$$ \tag{**}
 f = c g \text{ for some } c > 0 \, .
$$
Conversely, if $(**)$ holds for non-zero $f, g \in V$ then
$$
\langle f, g\rangle = c \langle g, g\rangle = c \| g \|^2 = \| f \|\| g \|
$$
so that $(*)$ and $(**)$ are actually equivalent.
If $V$ is the space $L_2(I)$ of square-integrable functions on some interval $I$ with the inner product
$$
\langle f, g\rangle = \int_I f(x) g(x) \, dx
$$
then functions which agree almost everywhere are identified.
In that case $f=cg$ in $L_2$ means that $f(x) =cg(x)$ a.e. in $I$.
