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Let $M\in\mathbb{R}^{n\times n}$ and $y\in\mathbb{R}^n$. We define $l:\,\mathbb{R}^n\to\mathbb{R}$ by $x\mapsto ||Mx-y||^2_2$. Now I´m looking for a a function $f:\,\mathbb{R}_{\geq 0}\to\mathbb{R}$, which satisfies $f(||x||_2)\leq l(x)$. In addition $f$ should be monotone increasing with $\lim_{\alpha\to\infty}\frac{f(\alpha)}{\alpha}=\infty$. This looks very simple and maybe it is, but unfornately I couldn't find such a function.

Edit: We assume $M\neq0$. If it helps we can also assume that $M$ is a diagonal binary matrix.

Thank you in advance,

Chris

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  • $\begingroup$ I updated my answer to reflect the edit. $\endgroup$ – user159517 Mar 17 at 14:01
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Let $M$ be a diagonal binary matrix. There are two possible cases to consider

  1. $M$ is singular
  2. $M$ is the identity matrix.

First, lets assume that 1) holds. Let $x_0 \in \text{ker} M$ with $\|x_0\| = 1$. Let $\alpha$ be big enough such that $f(\alpha) > \|y\|_2^2$. Then for all $c > \alpha$, we find

$$f(\|cx_0\|_2) > l(cx_0).$$ So the statement is wrong.

In the case $2)$, where $M$ is the identity matrix, take $$g(\|x\|_2) := (\|x\|_2- (1+\|y\|_2))^{+} $$ where $+$ denotes the positive part. Whenever $g$ is nonzero, due to the reverse triangle inequality we have $\|x-y\|_2 \geq 1$ and $$g(\|x\|_2) \leq \|x\|_2 - \|y\|_2 \leq \|x-y\|_2.$$ Now for any $p \in (1,2)$, the function $$f(\|x\|_2) := g(\|x\|_2)^p$$ does the job.

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  • $\begingroup$ That's right. I did not think about that corner case. I assume that $M\neq 0$. I will edit the question. Thank you! $\endgroup$ – Chris S. Mar 17 at 12:48

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