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Is it valid to take $|- \infty | = \infty$?

or is the absolute value e.g. not defined for infinity?

Particularly,

if one wishes to argue that operator $f(x)=x$ is not bounded below on $\mathbb{R}_{-}$, then the definition for bounded belowness for operators says must be $\beta > 0$ s.t.

$$\| T x \| \geq \beta\|x\|$$

(and here the norm is abs)

but then if $x \rightarrow -\infty$?

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    $\begingroup$ Well it is true that $|x|\to\infty$ as $x\to -\infty$. Is this question just out of curiosity? What is the context? $\endgroup$ – Minus One-Twelfth Mar 22 at 19:32
  • $\begingroup$ You should be careful though. While @MinusOne-Twelfth is correct, the way you've written your title is not accurate at all, unless of course you are using a different number system. $\endgroup$ – Don Thousand Mar 22 at 20:01
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It is indeed true that $\lim_{x \to -\infty} |x| = \infty$ and can be elementary checked.

As for the operator note, it is indeed true, since either $-\infty$ or $\infty$ give an expression of the type $\infty \geq \beta \infty$ with $\beta >0$, thus the operator is unbounded.

Be careful when referring to boundedness of operators :

Let $T:X \to Y$ be a linear operator. A bounded linear operator is generally not a bounded function. Trully, in many cases one can find a sequence $x_{k} \in X$ such that $ \|Tx_{k}\|_{Y}\rightarrow \infty $. Instead, all that is required for the operator to be bounded is that : $$\frac{\|Tx\|}{\|x\|} \leq M < \infty$$ Thus, the expression $\|Tx\| \geq \beta \|x\|$ does not mean bounded belowness as we would say when talking about a common function.

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    $\begingroup$ The second part of your answer, where you address what the OP really cares about, is correct. I disagree about the first part. $\infty$ is not a number and $0-\infty$ or $|0-\infty|$ or "the distance to the point at $\infty$" make no sense. The absolute values that matter are applied to real numbers. $\endgroup$ – Ethan Bolker Mar 22 at 20:39
  • $\begingroup$ @EthanBolker True, this is why I said a limit statement shall be better, I think it's for the best to correct it, so thanks for pointing out too ! $\endgroup$ – Rebellos Mar 22 at 20:40
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Yes, that's correct. Note that in your example you may take $\beta=1$ since $\infty\ge 1\infty$.

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  • $\begingroup$ But does the bounded below definition then make sense? Intuitively $f(x)=x$ cannot be bounded below on $\mathbb{R}_{-}$. $\endgroup$ – mavavilj Mar 22 at 19:36
  • $\begingroup$ @mavavilj A real operator is defined as $T : D \to \mathbb R$ where $D \subseteq R$. If $D \equiv R$ then indeed $T(x)=x$ is unbounded anyway. $\endgroup$ – Rebellos Mar 22 at 19:39
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    $\begingroup$ But you say that take $\beta = 1$ as if you would make it bounded? But I think that $\infty \geq \beta \infty$ is an absurd statement, because infinities cannot be compared? Thus there does not exist lower bound. $\endgroup$ – mavavilj Mar 22 at 19:41
  • $\begingroup$ @mavavilj Since $\beta >0$, the inequality is indeed true, since $\beta \infty \equiv \infty$ and $\infty \geq \infty$ is true (the equality holds). $\endgroup$ – Rebellos Mar 22 at 19:43
  • $\begingroup$ Then how to show that $f(x)=x$ is unbounded operator? $\endgroup$ – mavavilj Mar 22 at 19:45
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Beware that in $n$ dimensions (as the use of the norm seems to indicate), $x\to\infty$ and $x\to-\infty$ are meanigless/useless, because vectors can escape to infinity in many different ways. Such as $(t,-t,3,-\sin t,-t)$ where $t$ tends to infinity.

So all you consider is the norm, a scalar, that is always positive.

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