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Hello everyone I have the following question:

Let $\{x_k\} \in \mathbb{R}^n$ be a sequence such that $\|x_k\|^q_q \leq M$ for every $k \in \mathbb{N}$, where $\|x\|^q_q = \sum^n_{i=1}|x_i|^q$ and $M>0$. I know that $\|\cdot||_q$ it's not a norm and it doesn't even induces a metric as mentioned here (L1 convergence and Lp bounded implies Lq convergence). But I was reading an article which says if $\|x_k\|^q_q \leq M$ then $\{x_k\} $ is bounded and it has a subsequence which is convergent. But we say that a set $X$ in a metric space is bounded when for every $x,y \in X$ we have that $d(x,y) \leq M, \, M>0 $. How can a pseudo norm bound a sequence?

Thank you

Kutz
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    $d(x,y) := \sum_1^n |x_i - y_i|^q$ (note carefully the powers, this no longer scales like a norm) is a metric for $q\in(0,1)$ – Calvin Khor Sep 17 '19 at 01:19
  • You switch between using a subscript, to using a superscript, to not having anything, are they all the same? – Calvin Khor Sep 17 '19 at 02:09
  • @CalvinKhor In order to prove that is metric, I have to prove $d(x,y) = |x - y|^q \leq |x-z|^q + |z-y|^q$ for $0<q<1$. I see that the following result is true: let $a, b \geq 0$ then $a^q + b^q \leq (a + b)^q$ but I can't prove this result yet. Finally if d(x,y) = \sum |x_i - y_i|^q is a metric a think the boudendness of sequence comes easily(by definition) – Kutz Sep 17 '19 at 11:45
  • Well here's the proof. https://math.stackexchange.com/questions/295551/concave-implies-subadditive/2676802#2676802 Please also read Kavi's answer, and if it doesn't already answer your question, you should add this 'article' to your question because its not clear to me what you want – Calvin Khor Sep 17 '19 at 11:47
  • @CalvinKhor Kavi already answered my question. Basicaly in a metric space with metric $d$ in order to show that a subset $X$ is bounded, we need to prove that for every $x,y \in X$ we have $d(x,y) \leq c$ where $c > 0$ is constant, and if we have a norm we can just prove that $||x || \leq c$. What about a pseud norm? We don't the same, at least i think! Argumentation below just showed that boudedness by a specific pseudo norm implies in boundedness by norm. By the way, thank for your attention – Kutz Sep 17 '19 at 23:19

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In this argument boundedness is not w.r.t the given pseudo-metric. Since $|x_i|^{q} \leq M$ for each $i$ the sequence is bounded in the usual sense and hence it has a convergent subsquence. You can see that this convergence takes place w.r.t. the given pseudo-metric also.

Kavi Rama Murthy
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  • Since $|x_i|^q \leq M$ implies that $|x_i| \leq M^{1/q}$ for every $i \in {1, \ldots, n}$ so $ x = (x_1,\ldots, x_n) $ is bounded w.r.t $L_1$ norm then $x$ is bounded. is it your point?

    By the way, when you say usual sense, are you saying that $X$ is bounded when for every $x, y \in X$ we have $d(x,y) \leq c$ where $c > 0$ and $d$ is a metric?

     – Kutz Sep 17 '19 at 11:54
  • Yes, exactly. @Kutz – Kavi Rama Murthy Sep 17 '19 at 11:56