# Is the bilinear form $A(\cdot,\cdot)$ $l_{2}$-elliptic or coercive?

Consider the space $$l_{2} = \{ (x_{n})_{n\in \mathbb{N}} \subset \mathbb{R} : \sum_{n=1}^{\infty} |x_{n}|^{2} < \infty \}$$. Now, consider the bilinear form $$A(\cdot,\cdot):l_{2}\times l_{2}\to \mathbb{R}$$, which is given by $$$$A(x,y) = \sum_{n=1}^{\infty} a_{n}x_{n}y_{n}\text{,}$$$$ where $$(a_{n})_{n\in\mathbb{N}}$$ is a sequence of positive numbers defined by $$a_{n} = 2^{-n}$$. We know that $$l_{2}$$ is an inner product space with inner product $$\langle \cdot, \cdot \rangle$$ defined by $$$$\langle x,y \rangle = \sum_{n=1}^{\infty} x_{n}y_{n}\text{.}$$$$ Question: Is the bilinear form $$A(x,y)$$ coercive? In other words, does there exists $$\alpha>0$$ such that $$$$\sum_{n=1}^{\infty} a_{n}x_{n}^2 = A(x,x) \geq \alpha \sum_{n=1}^{\infty} x_{n}^{2}?$$$$

• If $\sum a_{n}x_{n}^2 \geq \sum \alpha x_{n}^2$ then $\sum ( a_{n} - \alpha) x_{n}^2\geq 0$. Since $x_{n}^2\geq 0$ then $a_{n}-\alpha \geq 0$. Apr 1, 2023 at 20:24
• that's not true. $\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n^2} > 0$ but half the coefficients are negative. Try to choose a specific sequence $\{x_n\}$ that will give you $a_n\geq\alpha$ Apr 1, 2023 at 20:44
• @dezdichado Yeah, that's not true. Thank you. Do you think $A(\cdot,\cdot)$ is coercive? Apr 1, 2023 at 21:05
No, there is no $$\alpha > 0$$ such that $$\tag{*} \sum_{n=1}^{\infty} 2^{-n}x_{n}^2 \geq \alpha \sum_{n=1}^{\infty} x_{n}^{2}$$ for all $$x = (x_n) \in \ell_2$$. If we fix a positive integer $$m$$ and choose for $$x$$ the sequence $$(\delta_{m, n})_n$$ which is $$1$$ at position $$m$$, and $$0$$ otherwise, then $$(*)$$ gives $$2^{-m} \ge \alpha > 0$$ for all positive integers $$m$$, which is not possible.
A similar argument shows that, given $$\alpha > 0$$, $$\sum_{n=1}^{\infty}a_n x_{n}^2 \geq \alpha \sum_{n=1}^{\infty} x_{n}^{2}$$ holds for all $$x = (x_n) \in \ell_2$$ if and only if $$a_n \ge \alpha$$ for all $$n$$. In other words, $$A(x,y) = \sum_{n=1}^{\infty} a_{n}x_{n}y_{n}$$ is coercive if and only if $$\inf \{ a_n \mid n \in \Bbb N \} > 0$$.