am trying to follow the outline of a proof in a book i am reading - must be missing something obvious, but would like to understand what exactly...
$f$ is complex and square integrable over e. g. [0, 1]: $\int_0^1 \lvert f(x) \rvert ^ 2 dx < \infty$
the inner product $<f,f> = \lVert f\rVert ^2 = \int_0^1f(x)\overline{f(x)} dx = \int_0^1 \lvert f(x) \rvert ^ 2 dx < \infty$ and $\lVert f \rVert = \sqrt{ {\lVert f \rVert}^2 }$
now, the book says that absolute integrability would follow with cauchy-schwarz:
$\lvert<v, w>\rvert \leq \lVert v \rVert \cdot \lVert w \rVert$
and just picking $g(x) = 1$ as the second function. so, i'm doing
$\lvert <f, g> \rvert = \lvert \int_0^1 f(x) dx \rvert \leq \lVert f \rVert$
after playing with the root, it seems that indeed, since $\lVert f\rVert ^2 < \infty$ also $\lVert f\rVert < \infty$
but this just gives me an upper bound for $\lvert \int_0^1 f(x) dx \rvert$, not for $\int_0^1 \lvert f(x) \rvert dx$, which confuses me, since
$\lvert \int_0^1 f(x) dx \rvert \leq \int_0^1 \lvert f(x) \rvert dx$.
i think what i want to arrive at is exactly https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality for the case $ p=q=2$...
my question now is: what am i getting wrong? what am i missing?
You can let $g$ be such that $f(x)\overline{g(x)} = |f(x)|$ and $|g(x)|=1$ for all $x \in [0,1]$. $\overline{g}=|f|/f$ on the set $\{x \in [0,1]:|f(x)|>0 \}$, so $g$ can be chosen measurable. Then by the Cauchy-Schwarz inequality, $$ \int_0^1|f|dx = \left| \langle f,g \rangle \right| \leq \|f\|_2\|g\|_2 = \|f\|_2 < \infty. $$ Hence $f$ is Lebesgue-integrable.