For a given $M \colon \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $M$ is positive semidefinite, linear and self-adjoint operator, the seminorm $\left\lVert \cdot \right\rVert _{M}$ endowed by M is defined as: $$ \left\lVert x \right\rVert _{M} ^{2} = \left\langle Mx , x \right\rangle \geq 0 , \forall x \in \mathbb{R}^{n} . $$

My question is:

  1. if we know that $$ \left\lVert a \right\rVert \leq \left\lVert b \right\rVert + \left\lVert c \right\rVert $$ which gives us (by squaring both sides then apply) something like $$ \left\lVert a \right\rVert ^{2} \leq 2 \left\lVert b \right\rVert ^{2} + 2 \left\lVert c \right\rVert ^{2} $$ then can we conclude that $$ \left\lVert a \right\rVert _{M}^{2} \leq 2 \left\lVert b \right\rVert _{M}^{2} + 2 \left\lVert c \right\rVert _{M}^{2} $$ here $\left\lVert \cdot \right\rVert$ stands for the usual Euclidean norm.

  2. if there exists some $\mu > 0$ such that $$ \left\lVert M \right\rVert _{n,n} = \sup\limits_{\left\lVert x \right\rVert \neq 0} \dfrac{\left\lVert Mx \right\rVert}{\left\lVert x \right\rVert} \leq \mu < + \infty $$ then can we conclude that $$ \left\lVert x \right\rVert _{M} ^{2} \leq \mu \left\lVert x \right\rVert ^{2} , \forall x \in \mathbb{R}^{n} . $$ I tried to take the square in the definition of the operator norm but it only leads to nowhere as I only have $$ \left\lVert Mx \right\rVert ^{2} \leq \mu^{2} \left\lVert x \right\rVert ^{2} , \forall x \in \mathbb{R}^{n} . $$

  1. No. Take for example $M:\mathbb{R}^3\to\mathbb{R}^3$ as $M(x,y,z)=(x,0,0)$. $M$ is linear, positive semidefinite and self-adjoint. If we consider $a=(1,0,0),b=(0,1,0),c=(0,0,1)$ we get:




  1. Yes. In fact for all $x$ we have $\|Mx\|\le\mu\|x\|$ and then:

$\|x\|_M^2=\langle Mx,x\rangle\le\|Mx\|\|x\|\le\mu\|x\|^2,$

where the first inequality is Cauchy-Schwarz inequality.

Also: note that in question 2 the hypothesis of the existence of $\mu>0$ is always verified (if $M\neq0$), in fact it is equivalent to the hypothesis that the operator $M$ is continuous, which is always true in finite dimensional spaces (for example see "Every linear mapping on a finite dimensional space is continuous").

  • $\begingroup$ Many thanks. I know where did I miss in 2. About 1 is there any way to archive some inequality for the semi norm, not the one I wrote but maybe different inequality? as I need to deduce the semi norm using the norm $\endgroup$ – mortal Sep 18 '17 at 11:48
  • 1
    $\begingroup$ @khonglagica To be honest I don't know if seminorms can get the same inequalities satisfied by the norm in general. I think that without further hypotheses it is impossible to give a general answer. $\endgroup$ – Pozz Sep 18 '17 at 12:46

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