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

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 $$x \rightarrow -\infty$$?

• Well it is true that $|x|\to\infty$ as $x\to -\infty$. Is this question just out of curiosity? What is the context? – Minus One-Twelfth Mar 22 at 19:32
• 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. – Don Thousand Mar 22 at 20:01

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.

• 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. – Ethan Bolker Mar 22 at 20:39
• @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 ! – Rebellos Mar 22 at 20:40

Yes, that's correct. Note that in your example you may take $$\beta=1$$ since $$\infty\ge 1\infty$$.

• But does the bounded below definition then make sense? Intuitively $f(x)=x$ cannot be bounded below on $\mathbb{R}_{-}$. – mavavilj Mar 22 at 19:36
• @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. – Rebellos Mar 22 at 19:39
• 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. – mavavilj Mar 22 at 19:41
• @mavavilj Since $\beta >0$, the inequality is indeed true, since $\beta \infty \equiv \infty$ and $\infty \geq \infty$ is true (the equality holds). – Rebellos Mar 22 at 19:43
• Then how to show that $f(x)=x$ is unbounded operator? – mavavilj Mar 22 at 19:45

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.