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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.

ananta
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tnt235711
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  • what is the reasoning for your argument ? – dezdichado Apr 01 '23 at 19:47
  • 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 $a_{n}-\alpha \geq 0$. – tnt235711 Apr 01 '23 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 $a_n\geq\alpha$ – dezdichado Apr 01 '23 at 20:44
  • @dezdichado Yeah, that's not true. Thank you. Do you think $A(\cdot,\cdot)$ is coercive? – tnt235711 Apr 01 '23 at 21:05

1 Answers1

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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 $\inf \{ a_n \mid n \in \Bbb N \} > 0$.

Martin R
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