For my Fourier analysis class, we have to prove this identity for piecewise continuous $f$ on an interval $(\alpha,\beta)$:
$$\int_\alpha^\beta | f(t) | dt \leq \sqrt{\beta - \alpha} \cdot || f ||$$
We are advised to make use of the Cauchy-Schwarz inequality for inner products (stated as $| \left \langle f,g \right \rangle | \leq ||f|| \cdot ||g||$) and the fact $f = f \cdot 1$.
Initially, working with $g(t)=1$ suffices to yield the following (skipping over a few formalities of the definition and calculation of inner product and norm for brevity's sake):
$$| \left \langle f,1 \right \rangle | = \left | \int_\alpha^\beta f(t) dt \right |$$
$$||1|| = \int_\alpha^\beta dt = \sqrt{\beta - \alpha}$$
Thus, so far by my calculations and the Cauchy-Schwarz inequality,
$$| \left \langle f,1 \right \rangle | = \left | \int_\alpha^\beta f(t) dt \right | \leq \sqrt{\beta - \alpha} \cdot ||f|| = ||f|| \cdot ||1||$$
Of course, this isn't quite the identity we want; that is, to reiterate
$$\int_\alpha^\beta | f(t) | dt \leq \sqrt{\beta - \alpha} \cdot || f ||$$
(The difference being that the absolute value needs to be moved into the integral.)
I imagine I need to make use of the identity
$$\left | \int_\alpha^\beta f(t) dt \right | \leq \int_\alpha^\beta | f(t) | dt$$
yet it's probably a little more nuanced than just applying it directly. (We'd essentially have a case of where we're saying $1 \leq 2$ and $1 \leq 4$ - doesn't mean $4 \leq 2$. In this case, $1$ being the absolute value of the integral, $2$ being the root and norm, and $4$ being the integral of the absolute value, as an oversimplification of why I don't think directly applying that identity would work.)
Does anyone have an idea of where to proceed from here? Thanks.
Since the question is still open and I can't accept a comment as an answer:
The solution followed from $⟨|f|,1⟩≤\sqrt{β−α}∥f∥$