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Is there an intuition for the infimum definition of ${\| A \|}_{\mathrm{op}}$ without using a different, equivalent definition?

I am referring to the definition, given an operator $A: W \rightarrow V$, $$ {\| A \|}_{\mathrm{op}} = \inf \{ c \geq 0 : \| A v \| \leq c \| v \|, \ \forall v \in V \}.$$

Note that the $\sup$ definition, $$ \| A \|_{\mathrm{op}} = \sup \{ \| A v \| : v \in V, \| v \| =1 \},$$ agrees more with my intuition and I know that one can prove they are equivalent (as I have done here), but that exercise seems conceptually unrevealing.

To me the $\inf$ definition reads as choosing the smallest $c$ that bounds all vectors, however my current intuition thinks that its a better question to think about how large $A$ makes things, so rather a maximum.

I guess trivially this means things are always bounded by infinity but my point is that it doesn't really answer or correct my wrong intuition. I still feel the size of $A$ should be some sort of question about how it makes things large, so why do we consider the smallest?

Joseph Martin
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Charlie Parker
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  • I don't know what else you are asking for; the inf in your first formula equals the sup in the second, so they are simply two ways of writing the same thing. – Giuseppe Negro Mar 18 '19 at 21:46
  • If you want $A$ to be bounded (equivalently continuous) then you do not want $A$ to make things too "large", which is why we take an infimum. Another way to think of $||A||_{op}$ is the smallest constant for which $A$ is Lipschitz continuous. – vxnture Mar 18 '19 at 21:47
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    @GiuseppeNegro I don't want proves. What I am looking for is if there is a way to explain/argue why ||||≤|||| for all ∈ is a good definition directly without referencing other stuff? Somehow the minimum is a good definition. Why? Can that be justified directly? – Charlie Parker Mar 18 '19 at 21:47
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    @nammie right, I am asking exactly why the Liptschitz continuous definition makes sense directly rather than appealing to equivalent definitions that seem to stand conceptually on their own. – Charlie Parker Mar 18 '19 at 21:49
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    You can find other textbooks, where your sup formula is used as the definition, and it is then proved to be equal to the inf formula. – GEdgar Mar 18 '19 at 21:51
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    @GEdgar I know the proves. Thats not what i am looking for. I am looking for a conceptual justification for the infimum definition on its own. – Charlie Parker Mar 18 '19 at 21:52
  • @Pinocchio One way this definition is useful is when you want to compute the value of $||A||{op}$. Say you know that $||A||{op} \geq c$. If you can construct a sequence $(v_n)$ so that $||A(v_n)|| \leq (c + 1/n)||v_n||$, then you have $||A||_{op} = c$. – vxnture Mar 18 '19 at 22:01

3 Answers3

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In the first formulation, the $c$ gives an upper bound to how much the vector may grow (that's what $\le$ means). Taking the infimum then says that you want the lowest upper bound to the vector's growth.

Now remember what the supremum is: The supremum of a set is, by definition, the lowest upper bound to that set. In other words, by asking for the lowest upper bound, you are really asking for the supremum of the vector's growth.

Indeed, you can make this quite explicit in the formulas:

Start with the definition you gave:

$$\| A \|_{\rm op} = \inf \{ c \geq 0 : \| A v\| \leq c \|v\|, \forall v \in V \}$$

Now we observe that the zero vector doesn't constrain $c$ in any way. Therefore let's remove that from the considered set.

$$\| A \|_{\rm op} = \inf \{ c \geq 0 : \| A v\| \leq c \|v\|, \forall v \in V\setminus\{0\} \}$$

Now that we have excluded the zero vector, we see that $\|v\| > 0$, therefore we can divide the inequality by it without changing anything. This gives:

$$\| A \|_{\rm op} = \inf \left\{ c \geq 0 : \frac{\| A v\|}{\|v\|} \leq c , \forall v \in V\setminus\{0\} \right\}$$

But now the right hand side explicitly describes the lowest upper bound to $\|Av\|/\|v\|$, that is,

$$\| A \|_{\rm op} = \sup \left\{ \frac{\| A v\|}{\|v\|} : v \in V\setminus\{0\} \right\}$$

And here you've got your supremum.

Note again that in the last step, I did nothing but literally apply the definition of the supremum.

celtschk
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The definition with the $\inf$ resembles the one used for constructing the Minkowski functional which, in turn, plays a key role in the proof of the Hahn-Banach theorem. The definition of the functional is carried out without using a norm, hence has meaning in any real vector space, normed or not. It is the use of the $\inf$ construct that allows to define the functional without the norm.

See, for example, Section 4.14 in Introductory Real Analysis, or the first chapter in Brezis's textbook on functional analysis, or Section 3.8 in Fundamentals of Functional Analysis.

avs
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An informal, but hopefully intuitive, answer:

Write $\partial B_{r}$ for the set $\{ v \in V \ | \ \| v \| = r \}$. Think of the dotted hexagon as being $A(\partial B_{1})$ (the result of applying $A$ to $\partial B_{1}$), and think of the blue circle as being $\partial B_{{\| A \|}_{\mathrm{op}}}$.

$\hspace{6cm}$enter image description here

To find ${\| A \|}_{\mathrm{op}}$, you can either find the biggest value of $\| A v \|$ for $v \in \partial B_{1}$, or the smallest $c$ such that $\partial B_{c} \supseteq A(\partial B_{1})$. The $\sup$ definition is the first method, and the $\inf$ definition is the second method.

The $\inf$ definition has $\partial B$ start out at infinity and shrink down to the smallest $c \geq 0$ so that $\partial B_{c}$ still contains $A(\partial B_{1})$, which is exactly the same as the smallest $c \geq 0$ so that $\| A v \| \leq c \| v \|$ $\forall v \in V$.

Joseph Martin
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