My favorite interpretation of the identity
$$
\langle x,y \rangle = \|x\| \|y\| \cos \theta \tag{1}
$$
is that (1) defines the angle $\theta \in [-\pi/2,\pi/2]$ between the vectors $x$ and $y$. Since
$$
\langle x,y \rangle =\sum_{j=1}^n x_j y_j
$$
is the standard inner product in $\mathbb{R}^n$, the elementary Cauchy-Schwarz inequality implies that
$$
-1 \leq \frac{\langle x,y \rangle}{\|x\| \|y\|} \leq 1,
$$
and therefore there exists a unique angle $\theta \in \left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$ such that (1) holds.
However, once you have fixed $x$ and $y$, the question becomes two-dimensional in the plane spanned by $x$ and $y$. There you can apply stadard elementary geometry to show that the cosine of the angle between the two vectors is given by (1).
I think the space $W$ should be defined a partial order $\leq$ and zero element $0$ firstly and satisfy:
- If $a\leq b$ then $ca\leq cb$, $\forall a,b\in W$ and $0\leq c\in W$.
- If $a\leq b$ then $a-b\leq 0$.
Secondly, a multiply operator $\cdot$ should be defined in $W$ and satisfy $0\leq a\cdot a\stackrel{\triangle}{=}a^2$ for $\forall a\in W$. Also, the inverse operator of $\cdot$ should be defined in $W$ (Alternatively, the inverse element is defined in $W$). That is, if $ab=c$ then $c\stackrel{\triangle}{=}a/b$ for $\forall a,b,c\in W$ and $b\neq 0$ where $/$is the inverse operator of $\cdot$. What is more, these operators should be closed in $W$. Say, if $\forall a,b\in W$ then $a\cdot b\in W$ and $a/b\in W$ if $b\neq 0$. Finally, the operators $\cdot$ and /should satisfy commutative law.
Thirdly, there should have a multiply operator between the elements from $W$ and $V$ because we will define inner product by using this operator.
What is more, to hold the Cauchy-Schwarz inequality, the properties of inner product is important. I believe the Cauchy-Schwarz inequality is valid in a space which define a inner product whose definition is classical. In another word, if a space $V$ have been defined a inner product $(*,*)$ (say, a bilinear form that $V\times V\rightarrow W$) satisfy the following conditions:
- Commutative: $(x,y)=(y,x)$, $\forall x,y\in V$ (If V is a complex space, the right hand side should be dual. But for the sake of simplicity, we ignore it here.)
- Linearity: $(\alpha x+\beta y, z)=\alpha(x,z)+\beta(y,z)$, $\forall x,y,z\in V$ and $\alpha,\beta\in W$.
- Positive define: $(x,x)\geq0$, $\forall x\in V$. The equal sign is valid iff $x=0$ is valid where $0$ donate the zero element in $W$.
Then, by this definition, the Cauchy-Schwarz inequality is valid. The proof are as follow:
For $\forall\lambda\in W$ and $\forall x,y \in V$, we have:
\begin{equation}
0\leq (x+\lambda y,x+\lambda y)=(x,x)+2\lambda(x,y)+\lambda^2(y,y)
\end{equation}
If $y=0$, that is a trivial case and Cauchy-Schwarz inequality is valid obviously. If $y\neq 0$, let $\lambda=-(x,y)/(y,y)$ then we have:
\begin{equation}
0\leq(x,x)-2(x,y)^2/(y,y)+(x,y)^2/(y,y)^2(y,y)\\
(x,y)^2\leq (x,x)(y,y)
\end{equation}
This is the Cauchy-Schwarz inequality.
In fact, Cauchy-Schwarz inequality imply that the inner product of two elements is less than the their product of length because there is an angle between them. And $W$ is a space to measure the inner product of $V$. So I think the conditions I assume at start is reasonable.
Best Answer
You can use the 'scalar' version of the Cauchy-Schwarz inequality more or less directly:
You have $f^T g \le \|f\| \|g\|$ for any $f,g \in \mathbb{R}^n$.
Hence $|\int f^T(t) g(t) dt |\le \int \|f(t)\| \|g(t)\| dt \le \sqrt{\int \|f(t)\|^2dt} \sqrt{\int \|g(t)\|^2dt}$