I learned in my abstract linear algebra class that suppose $u,v\in V$, then $$|\langle u,v\rangle| \le \|u\|\|v\| .$$ The way I always remember is that the dot product of two vectors is less or equal than the product of their norms.
In the case that $V$ is a real vector space of continuous real-valued functions on the interval $[-1,1]$, the inner product of $f,g$ is $$\langle f, g \rangle = \int_{-1}^1 fg dx$$ and Cauchy-Schwartz inequality implies $$|\langle f,g\rangle|^2=(\int_{-1}^1 fg~ dx)^2 \le \int_{-1}^1 f^2 dx\int g^2 dx = \|f\|^2 \|g\|^2,$$ which aligns with the above general statement.
However, consider the following case. $(X,m,\mu)$ is a measure space. Let $E\in \mathfrak{M}$ and let $f:X\to \mathbb{C}$. And set $L^2(E,\mu)$ with distance function 2-norm. Let $f\in L^2(E,\mu), g \in L^2(E,\mu)$. The Cauchy-inequality given by Rudin on page 326 is $$ \int_X |fg| d\mu\le \|f\|\|g\|$$.
I'm confused about the left-hand side. Because $\int_X |fg| d\mu$ does seems to be the inner product $\langle f,g \rangle = \int f\bar g ~d\mu$.
Is it correct to think that the Cauchy-Schwartz inequality in the last case is not exactly aligned with the general statement mentioned in the very beginning?
Best Answer
The identity given by Rudin is $\|fg\|_1 \leq \|f\|_2 \|g\|_2$ which is equivalent to Cauchy's inequality:
$\implies\quad$ $ |\langle f, g \rangle| = \left| \int f \bar{g} \right| \leq \int |f \bar{g}| = \int |fg| = \|fg\|_1 \leq \|f\|_2 \|g\|_2. $
$\impliedby\quad$ $ \|fg\|_1 = \int |fg| = \int |f||g| = \langle |f|, |g| \rangle \leq \| |f| \|_2 \, \| |g| \|_2 = \|f\|_2 \|g\|_2. $