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

Question

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 $x \rightarrow -\infty$?

Answer 1

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

Answer 2

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.

Answer 3

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.