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
\begin{equation}
A(x,y) = \sum_{n=1}^{\infty} a_{n}x_{n}y_{n}\text{,}
\end{equation}
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
\begin{equation}
\langle x,y \rangle = \sum_{n=1}^{\infty} x_{n}y_{n}\text{.}
\end{equation}
Question: Is the bilinear form $A(x,y)$ coercive? In other words, does there exists $\alpha>0$ such that
\begin{equation}
\sum_{n=1}^{\infty} a_{n}x_{n}^2 = A(x,x) \geq \alpha \sum_{n=1}^{\infty} x_{n}^{2}?
\end{equation}
Thanks in advance.
Best Answer
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 $\inf \{ a_n \mid n \in \Bbb N \} > 0$.