Inverse of Cauchy Schwarz inequality

Question

Assume a finite real number sequence $a_i, i=1,\ldots,N$. The following inequality holds true: \begin{align*} \sum_{i=1}^Na_i^2\leq\left(\sum_{i=1}^N\lvert{a_i}\rvert\right)^2\leq N\sum_{i=1}^Na_i^2. \end{align*} The RHS inequality is the well-known Cauchy Schwarz inequality. Can the LHS inequality be generalized to Hilbert space like the Cauchy Schwarz inequality? Thanks.

Answer 1

This isn't true in general Hilbert spaces; the LHS inequality would be akin to $\langle x, x \rangle \le \langle x, 1\rangle^2$. Take $H = C^0$ (set of continuous functions) with inner-product $\langle f, g \rangle = \int_{1}^{2} f(x) g(x) dx$ and let $f(x) = x$ for a counterexample.

Answer 2

Suppose f and g are real-valued positive functions. Is there any inequality like $\langle f,g\rangle\geqslant C\|f\|\|g\|$ for some $C$?

I think it is still wrong even we require $f$ and $g$ are strictly positive ($f,g > 0$).

Let's consider the real-valued function on closed interval. Such as $[0,1]$. With the linearity of the inner product and norm, it is equivalent to consider the functions with norm $1$ ($\|f\|_{L_2}=1$). Thus the RHS is $C$. It is easy to construct $f$ and $g$, and their inner product $\langle f,g\rangle < C$.