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$$\|A\|_1=\inf\{c\ge 0: \forall x\in V,\|Ax\|\le c, \|x\| \le 1\}=\inf\ S_1$$ $$\|A\|_2=\inf\{c\ge 0: \forall x\in V,\|Ax\|\le c, \|x\| = 1\}=\inf S_2$$ I am trying to prove these two expresions are equal DIRECTLY, I know there are other equivalente definitions using the sup, I know how to deal with those, so let's not use them.

Similarly to what is done to prove the equivalence between the sup version of these two expresions ( here Why are different definitions of the operator norm equivalent?), I notice the inclusion between sets: $S_2 \subset S_1$ , so

$\|A\|_1=\inf\ S_1 \le \inf\ S_2 =\|A\|_2$ ...(*)

If I can prove the opposite inequality I'd be done, that is where I am having trouble.

I know that for $ x \in V, \|x\|\le 1 , \|Ax\| \le \|Ax\|/\|x\| $. I don't want to take the sup over $ x$ here since that is the reasoning for the sup version of this problem, so how do I go about reasoning in terms of the inf of the constants that bound each side of the inequality? Let $C_1$ represent any value of $C$ in $S_1$ and $C_2$ represent any value of $C $ in $S_2 $.

I know $\|Ax\| \le \|A\|_1 \le C_1$ $\forall \|x\| \le 1$

and $\|Ax\| \le \|A\|_2 \le C_2$ $\forall \|x\| = 1$ that is $ \|Ax\|/\|x\| \le \|A\|_2 \le C_2 \forall x \in V$

How can I conclude $\|A\|_2\le \|A\|_1$ ?

some_math_guy
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2 Answers2

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Let us do it very formally. We write $D = \{ x \in V \mid \lVert x \rVert \le 1 \}$ and $B = \{ x \in V \mid \lVert x \rVert = 1 \}$. Your sets $S_1, S_2$ are then given as

$$S_1 = \{ c \ge 0 \mid \forall x \in D : \lVert Ax \rVert \le c \}, \\ S_2 = \{ c \ge 0 \mid \forall x \in B : \lVert Ax \rVert \le c \} .$$ Note that for an infinite-dimensional $V$ these sets may be empty.

By definition we have $S_1 \subset S_2$ simply because $B \subset D$. In fact, if $\lVert Ax \rVert \le c$ for all $x \in D$, then trivially $\lVert Ax \rVert \le c$ for all $x \in B$.

Note that this argument is valid for arbitrary functions $A$, we did not use that $A$ is linear. Let us prove that for linear $A$ we have $S_2 \subset S_1$. This implies $S_1 = S_2$ and therefore $\lVert A \rVert_1 = \lVert A \rVert_2$.

So let $c \in S_2$ and $\lVert x \rVert \le 1$. If $\lVert x \rVert = 0$, then $x = 0$ and $\lVert Ax \rVert = 0 \le c$. If $\lVert x \rVert \ne 0$, then $y = x/\lVert x \rVert$ is defined and $\lVert y \rVert = 1$. Thus $\lVert Ay \rVert \le c$. But we have $$\lVert Ax \rVert = \lVert A(\lVert x \rVert \cdot y) \rVert = \lVert \lVert x \rVert Ay \rVert = \lVert x \rVert \cdot \lVert Ax \rVert \le 1 \cdot c = c .$$

Paul Frost
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  • Does the following argument work? Since $B \subset D$, In $S_1$ we have all the uper bounds $c\ge 0 $that bound $|Ax|$ for $|x| < 1$ in addition to the upper bounds $c\ge 0 $ that bound $|Ax|$ for $ |x| = 1$, which are the only ones in $S_2$. So the set $S_1$ has no less elements than and includes those of $S_2$, thus$|S_2| \le |S_1|$ and $ S_2\subset S_1$ – some_math_guy Jan 31 '22 at 12:06
  • @J.C.VegaO No. Your argument does not use the linearity of $A$. But for non-linear functions it is definitely false. – Paul Frost Jan 31 '22 at 12:10
  • But why doesn't it work? And I am not very convinced that "$S_1\subset S_2$ simply because $B\subset D$". It seems to regard $S_1$ and $S_2$ as subsets of V: $S_1={x |....}$, $S_2={x |....}$ instead of as subsets of numbers $S_1={c \ge 0 |....}$, $S_2={c \ge 0 |....}$ – some_math_guy Jan 31 '22 at 12:16
  • This question is in the context of equivalent definitions for the norm of a linear bounded operator – some_math_guy Jan 31 '22 at 12:18
  • What do you want to prove with the argument in your first comment? $S_1 \subset S_2$ or $S_2 \subset S_1$ as in your question ("I notice the inclusion")? My argument for $S_1 \subset S_2$ is based on the fact that $ \forall x \in D : \lVert Ax \rVert \le c$ implies $ \forall x \in B : \lVert Ax \rVert \le c$ simply because $B \subset D$. Now if $c \in S_1$, then it has the property $ \forall x \in D : \lVert Ax \rVert \le c$ , thus also the property $ \forall x \in B : \lVert Ax \rVert \le c$ which means $c \in S_2$. – Paul Frost Jan 31 '22 at 13:36
  • And $S_2 \subset S_1$ as you say in your comment is false if $A$ is not linear. Consider $A : \mathbb R \to \mathbb R, A(x) = 1 - \lvert x \rvert$. Then $S_1 = [1,\infty)$ and $S_2 = [0,\infty)$. Since your argument does not use any property of $A$, it cannot prove that $S_2 \subset S_1$. – Paul Frost Jan 31 '22 at 13:42
  • With my first comment I was trying to prove $S_2 \subset S_1$, as in my question. I guess I should have specify A is a linear and bounded operator. I don't know about non linear operators, so I didn't know the same definition applied to them. So in the context of linear bounded operators, is it wrong anyway? – some_math_guy Jan 31 '22 at 14:45
  • @J.C.VegaO As I said, a proof of $S_2 \subset S_1$ must be based on the linearity of $A$ (I have used it in my answer). In your attempt you definitely do not use it, thus it does not work. – Paul Frost Jan 31 '22 at 16:06
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Since $\inf S_1 = \|A\|_1$ is an upper bound of $\|Ax\|$,

$\forall x ,\|x\| \le 1$ ,$\|Ax\| \le \|A\|_1 $.

In particular

$\forall x ,\|x\| = 1$, $\|Ax\| \le \|A\|_1 $,

so $\|A\|_1 \in S_2$, but since $\inf S_2 = \|A\|_2$ , it follows that $\|A\|_2 \le \|A\|_1$

some_math_guy
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  • To be formal, you need to justify why the $\inf$ can be replaced by $|A|_1$. – copper.hat Jan 26 '22 at 21:27
  • @copper.hat I don't follow you,$ \inf S_1=∥A∥_1$ is by assumption. The result here together together with (*) in the original post prove $∥A∥_2=∥A∥_1$ – some_math_guy Jan 26 '22 at 22:02
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    Well, it is terribly pedantic, but a priori you do not know that $\sup S_1$ is an upper bound of $|Ax|$ for $|x| \le 1$. You know that for any $\epsilon>0$ that $\sup S_1+ \epsilon$ is an upper bound and so $\sup_{|x| \le 1} |Ax| \le \sup S_1+ \epsilon$ for any $\epsilon>0$ hence $\sup_{|x| \le 1} |Ax| \le \sup S_1$. – copper.hat Jan 26 '22 at 22:02
  • @copper.hat Do you mean $\inf S_1$ instead of $\sup S_1$? Notice that the sup $S_1 = \infty$, since $S_1$ is the set of positive constants that bound $|Ax|$ – some_math_guy Jan 26 '22 at 22:12
  • My apologies, I mean $\inf S_1$, sorry. – copper.hat Jan 26 '22 at 22:13