Let $\mathbf{f}:A\to\mathbb{R}^{d}$ and $\mathbf{u}:B\to\mathbb{R}^{d}$ be two vector functions with coordinate functions $f_{j}$ and $u_{j}$, respectively. Then from the Cauchy-Schwarz inequality, it follows that:
\begin{equation}
\int_{t_{0}}^{t}\langle \mathbf{u}(t),\mathbf{f}(s)\rangle\,\mathrm{d}s\leq\int_{t_{1}}^{t_{2}}\lVert\mathbf{u}(t)\rVert\lVert\mathbf{f}(s)\rVert\,\mathrm{d}s, \tag 1
\end{equation}
where $t_{1}=\min\{t_{0},t\}$ and $t_{2}=\max\{t_{0},t\}$. Hence:
\begin{equation}
\left\lVert\int_{t_{0}}^{t}\mathbf{f}(s)\,\mathrm{d}s\right\rVert\leq\int_{t_{1}}^{t_{2}}\lVert\mathbf{f}(s)\rVert\,\mathrm{d}s. \tag 2
\end{equation}
My question: Why is it that the integral must have limits $t_{1}$ and $t_{2}$ after using the Cauchy-Schwarz inequality? And how does $(2)$ follow from $(1)$? Thanks in advance! Please let me know if there's something that does not make sense.
My thoughts: In order to use the Cauchy-Schwarz inequality in $(1)$ we need absolute values around the inner product in the integral such that:
\begin{equation}
\int_{t_{0}}^{t}\langle \mathbf{u}(t),\mathbf{f}(s)\rangle\,\mathrm{d}s=\int_{t_{0}}^{t} \lvert\langle \mathbf{u}(t),\mathbf{f}(s)\rangle\rvert\,\mathrm{d}s.
\end{equation}
However, might it be that the equality above is only true when $t_{0}<t$? If that is the case, then it would make sense why one defines $t_{1}$ and $t_{2}$ as above, right?
Best Answer
In the first instance, we may have $t < t_0$. Therefore, beginning with the inner product on the left of $(1)$, \begin{align} \int_{t_0}^t \langle \mathbf u(t), \mathbf f(s) \rangle ~ ds &\leqslant \left\lvert \int_{t_0}^t \langle \mathbf u(t), \mathbf f(s) \rangle ~ ds\right\rvert \\ &\leqslant \int_{t_1}^{t_2} \left\lvert \langle \mathbf u(t), \mathbf f(s) \rangle \right\rvert~ ds. \\ \end{align} In moving the modulus sign inside the integral we need the limits of integration to be positively oriented, hence the need to make the change so that $t_1 \leqslant t_2$. We can now apply Cauchy-Schwarz to the integrand, \begin{align} \int_{t_0}^t \langle \mathbf u(t), \mathbf f(s) \rangle ~ ds &\leqslant \int_{t_1}^{t_2} \lVert \mathbf u(t) \rVert ~\lVert \mathbf f(s) \rVert ~ ds. \end{align} Thus $(1)$ is obtained.
For $(2)$ take $\mathbf u(t)$ to be the unit vector defined by, $$ \mathbf u(t) = \frac{\int_{t_0}^t \mathbf f(x)~dx}{\lVert \int_{t_0}^t \mathbf f(x) ~dx \rVert}$$ and substitute it into $(1)$. The left hand side is seen to be simply \begin{align} \left \lVert \int_{t_0}^t \mathbf f(s)~ds \right \rVert \end{align} while the right side, because $\lVert \mathbf u \rVert = 1$, becomes, $$\int_{t_1}^{t_2} \lVert f(s) \rVert ~ ds, $$ and the required result follows.