We will identify \(\C\) with \(\R^2\text{.}\) Observe that under this identification, the modulus of a complex number \(z=a+ib\) is the norm of the vector \((a,b)\text{:}\) that is,
\begin{equation*}
\abs{a+ib}=\norm{(a,b)}=\sqrt{a^2+b^2}\text{.}
\end{equation*}
Under this identification, a complex-valued function \(f=f_1+if_2\) is identified as the vector-valued function
\begin{align*}
f\colon [a,b] \amp\rightarrow \R^2\\
t \amp \longmapsto (f_1(t),f_2(t))\text{,}
\end{align*}
and the integral \(\int_a^bf(t)\, dt\) is then identified as the vector-valued integral
\begin{equation*}
\int_a^bf(t)\, dt=\left(\int_a^bf_1(t)\, dt, \int_a^bf_2(t)\, dt\right)\text{.}
\end{equation*}
Lastly, for vectors \(\boldv=(a,b)\) and \(\boldw=(c,d)\text{,}\) we will denote their dot product as \(\angvec{\boldv, \boldw}\text{:}\) i.e.,
\begin{equation*}
\angvec{\boldv, \boldw}=ac+bd\text{.}
\end{equation*}
Since integration of vector-valued functions is linear, and since the inner product is linear in each variable, it follows easily that
\begin{equation}
\angvec{\int_a^bf(t)\, dt, \boldv}=\int_a^b \angvec{f(t), \boldv}\, dt\tag{1.48}
\end{equation}
for any integrable vector-valued function and vector \(\boldv\in \R^2\text{.}\)
In this setting we wish to show that
\begin{equation*}
\norm{\int_a^bf(t)\, dt}\leq \int \norm{f(t)}\, dt\text{.}
\end{equation*}
The claim is obvious if \(\int_a^b f(t)\, dt=\boldzero\text{.}\) Otherwise, since \(\norm{\boldv}^2=\angvec{\boldv,\boldv}\) for any \(\boldv\in \R^2\text{,}\) we have
\begin{align*}
\norm{\int_a^b f(t)\, dt}^2 \amp = \angvec{\int_a^b f(t)\, dt, \int_a^b f(t)\, dt} \\
\amp = \int_a^b \angvec{f(t), \int_a^b f(t)\, dt}\, dt \amp \knowl{./knowl/xref/eq_dot_int.html}{\text{(1.48)}} \\
\amp = \abs{\int_a^b \angvec{f(t), \int_a^b f(t)\, dt}\, dt} \amp (\text{quantity is nonnegative})\\
\amp\leq \int_a^b\abs{\angvec{f(t), \int_a^b f(t)\, dt}} \, dt \amp (\text{real int. prop.})\\
\amp \leq \leq \int_a^b\norm{f(t)}\norm{\int_a^b f(t)\, dt} \, dt \amp (\text{Cauchy-Schwarz ineq.}) \\
\amp = \norm{\int_a^b f(t)\, dt}\int_a^b\norm{f(t)}\, dt \text{.}
\end{align*}
In summary, we have
\begin{equation*}
\norm{\int_a^b f(t)\, dt}^2\leq \norm{\int_a^b f(t)\, dt}\int_a^b\norm{f(t)}\, dt\text{.}
\end{equation*}
Since we assume that \(\int_a^b f(t)\, dt\ne \boldzero\text{,}\) we have \(\norm{\int_a^b f(t)\, dt}\ne 0\text{,}\) allowing us to cancel and obtain
\begin{equation*}
\norm{\int_a^b f(t)\, dt}\leq \int_a^b\norm{f(t)}\, dt\text{,}
\end{equation*}
as desired.
