Given the Cauchy-Schwarz inequality and the Riemann definition for the integral, $$\sum_{k=1}^{n}a_kb_k\le\sqrt{\sum_{k=1}^{n}a_k^2}\sqrt{\sum_{k=1}^{n}b_k^2}$$ $$\int_a^bf(x)dx=\lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}f\left(a+k\frac{b-a}{n}\right)\frac{b-a}{n}\right)$$
why can you not simply put the two together $$\lim_{n\rightarrow \infty} \sum_{k=1}^{n}a_kb_k\le\sqrt{\lim_{n\rightarrow \infty}\sum_{k=1}^{n}a_k^2}\sqrt{\lim_{n\rightarrow \infty}\sum_{k=1}^{n}b_k^2}$$
$$a_k=f \left(a+k\frac{b-a}{n}\right)\sqrt{\frac{b-a}{n}},b_k=g \left(a+k\frac{b-a}{n}\right)\sqrt{\frac{b-a}{n}}$$ ($a_k,b_k$ depends on $n$) to get $$\int_a^bg(x)f(x)dx\le \sqrt{\int_a^bf(x)^2dx}\sqrt{\int_a^bg(x)^2dx}$$ ? It appears that this simple-minded method is incorrect, for example, here's a passage from page 11 of Michael Steele's book:
[This] approach to [the integral inequality] via Cauchy’s inequality would have been problematical for several reasons, including the fact that the strictness of a discrete inequality can be lost in the limiting passage to integrals.
Could some kind person explain why this is: in general and in this particular case (at which point did the above 'derivation' run afoul)?
The proof is right, the issue is the following:
If
$$\sum_{k=1}^{n}a_kb_k < \sqrt{\sum_{k=1}^{n}a_k^2}\sqrt{\sum_{k=1}^{n}b_k^2}$$
you only get
$$\lim_{n \to \infty} \sum_{k=1}^{n}a_kb_k\le\sqrt{\lim_{n\rightarrow \infty}\sum_{k=1}^{n}a_k^2}\sqrt{\lim_{n\rightarrow \infty}\sum_{k=1}^{n}b_k^2} $$
It is possible to have $c_n < d_n$ and $\lim c_n = \lim d_n$.
That paragraph refers to the fact that Schwartz needed to find/use the equality case in the CS.
So, you can extend the standard CS inequality by Riemann Sums, but you cannot characterize this way the equality case....