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I have a bilinear operator $B\colon X \times Y\to Z$ with $X,Y,Z$ normed spaces, and define a norm on $X \times Y$ by $\lVert(x,y)\rVert = \lVert x\rVert_X + \lVert y\rVert_Y$ (using the respective norms on $X,Y$). Does the definition of the operator norm then generalize to $\lVert B\rVert = \sup\{\lVert B(x,y)\rVert/\lVert(x,y)\rVert$, $(x,y) \neq 0\}$?

Reason for asking: assuming this, $B(x,y) = xy$ does not seem to be a bounded operator (from i.e. $\mathbb{R}^2\to\mathbb{R}$) in this norm (or euclidean norm/ max norm for that matter), which contradicts the continuity (or at least until now I always thought the product operator was continuous!).

edit: changed linear to bilinear

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Are you sure that the multiplication operator $B$ is linear? –  t.b. Jun 19 '11 at 20:02
    
Sorry, meant to say bilinear. And I think I just realised the issue, bounded <==> continuous only holds for linear operators, not bilinear ones? –  tobias Jun 19 '11 at 20:05
    
No, this holds for bilinear operators as well, but the correct norm is $\displaystyle \sup\limits_{\|x\|,\|y\|\leq1}\|B(x,y)\|$. –  t.b. Jun 19 '11 at 20:07
    
Thanks, so the product operator is bounded after all! Reassuring. –  tobias Jun 19 '11 at 20:10
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3 Answers

up vote 8 down vote accepted

To expand on my comment:$\DeclareMathOperator{\Hom}{Hom}$

There are many ways to equip $X \times Y$ with a norm, the most natural ones are $\|(x,y)\| = \max{\{\|x\|,\|y\|\}}$ and $\|(x,y)\| = \|x\| + \|y\|$ since they correspond to the categorical product and coproduct operations (in the category of Banach or normed linear spaces and linear maps of norm $\leq 1$). Be that as it may, it is a good exercise to check that all the $p$-norms $\|(x,y)\|_{p} = \left(\|x\|^{p} + \|y\|^{p}\right)^{1/p}$ on $X \times Y$ are equivalent, thus a linear map $T: X \times Y \to Z$ is continuous if and only if it is bounded with respect to any of the norms with which you can equip the space $X \times Y$.

Note that linear means $T(\lambda x, \lambda y) = \lambda T(x,y)$ and $T(x+x',y+y') = T(x,y) + T(x',y')$ for all $(x,y), (x',y') \in X \times Y$ and all $\lambda \in \mathbb{R}$.

The multiplication is not linear in this sense, but it is bilinear $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$. As such, it is obviously continuous and it is bounded with respect to the norm on the bilinear maps $B: X \times Y \to Z$ given by

$$\|B\| = \sup_{\|x\|,\|y\| \leq 1} \|B(x,y)\|_{Z}$$

and it is a good exercise to check:

  1. A bilinear map $B$ is continuous if and only if it is bounded with respect to the norm above. Note that this simply means that $\|B(x,y)\|_{Z} \leq \|B\| \,\|x\|_{X} \, \|y\|_{Z}$ by bilinearity.
  2. If $Z$ is a Banach space then the space of bilinear maps $X \times Y \to Z$ is complete with respect to that norm.

Added: In fact, this idea naturally leads to the notion of the projective tensor norm.

Recall that a bilinear map $B: X \times Y \to Z$ corresponds to a linear map $b: X \otimes Y \to Z$. An element of the tensor product can be written as a finite sum $\sum x_i \otimes y_i$ and since $b$ is linear we have $b(\sum x_i \otimes y_i) = \sum b(x_i \otimes y_i) = \sum B(x_i,y_i)$. Now we want a norm on $X \otimes Y$ such that $b$ is bounded if and only if $B$ is bounded. Now $B$ is bounded if and only if $B$ is bounded on the elementary tensors, for which we have $\|B(x,y)\| \leq \|B\|\,\|x\|\,\|y\|$ so the norm of $\sum x_i \otimes y_i$ on $X \otimes Y$ had better be comparable to $\sum \|x_i\| \, \|y_i\|$. Now the problem is that this is not well defined because an element of $X \otimes Y$ has many decompositions into sums of elementary tensors. It turns out that the correct definition for the norm of $\omega \in X \otimes Y$ is

$$\|\omega \|_{\pi} = \inf\left\{ \sum \|x_i\|_X \, \|y_i\|_{Y} \,:\, \omega = \sum x_i \otimes y_{i}\right\}$$

where the infimum is taken over all (finite) representations $\omega = \sum x_i \otimes y_i$ as sum of elementary tensors. It is quite obvious that $\|\cdot\|_{\pi}$ is a semi-norm on $X \otimes Y$ satisfying $\|x \otimes y\|_{\pi} \leq \|x\|_{X} \|y\|_{Y}$. A bit more work shows that actually $\|x \otimes y\|_{\pi} = \|x\|_{X} \|y\|_{Y}$ and that $\| \cdot \|_{\pi}$ is a norm. Moreover, one can check that $\|B\| = \|b\|$, when the latter is computed as operator norm on $X \otimes Y$ with respect to $\|\cdot \|_{\pi}$. This gives us a bijection $$\text{Bil} (X,Y;Z) = \text{Hom}(X \otimes Y, Z)$$ between the spaces of bounded bilinear maps $X \times Y \to Z$ and bounded linear maps $X \otimes Y \to Z$. Now if $X,Y$ happen to be Banach spaces then $X \otimes Y$ no longer is a Banach space in general, so we may simply complete it and we write $X \widehat{\otimes} Y$ for this completion. Combining this with the observation made by Mark in his answer, we get the (isometric) correspondences $$\Hom{(X \widehat{\otimes} Y, Z)} = \text{Bil}(X,Y;Z) = \Hom(X,\Hom(Y,Z))$$ which we know well from linear algebra. We equip all $\Hom$-spaces with their natural operator norms and the space $\text{Bil}$ with the norm I defined above.

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are the $p$-norms equivalent in the space $X\times Y$ even when $\dim X\times Y=\infty$? –  leo Jul 18 '11 at 22:57
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@leo: Yes, I'm saying that all the norms $$\|(x,y)\|_{p} = \left(\|x\|^{p} + \|y\|^{p}\right)^{1/p}$$ are equivalent on $X \times Y$, that's essentially because they are equivalent in $\mathbb{R}^2$. However, what you might have in mind is that the norms $$\|(x_n)\|_{p} = \left( \sum_{n=1}^{\infty} |x_n|^p \right)^{1/p}$$ are not equivalent on the space of infinite sequences, see here. –  t.b. Jul 18 '11 at 23:04
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@leo: Sorry, I didn't think that it was misleading, but reading it again, I that I stated it a bit confusingly. What I mean are two things: First, all the $l^{p}$-norms are equivalent on $\mathbb{R}^2$. That is to say, you have for $p$ and $q$ a constant $C \geq 1$ such that $$C^{-1} (|a|^p + |b|^{p})^{1/p} \leq (|a|^q + |b|^{q})^{1/q} \leq C (|a|^p + |b|^{p})^{1/p}$$ for all $(a,b) \in \mathbb{R}^2$. Now apply this to $a = \|x\|$ and $b = \|y\|$ to see that the $p$-norms on $X \times Y$ are equivalent ($X$, $Y$ are any two normed spaces). Of course, if $\dim{(X \times Y)} \neq 2$ then ... –  t.b. Jul 18 '11 at 23:24
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Thanks for the clarification, you are very kind. –  leo Jul 18 '11 at 23:30
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@leo: Thank you, that's nice to hear! Feel free to ask whenever something I say is unclear. See you around! –  t.b. Jul 18 '11 at 23:36
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A bilinear transformation $B: X \times Y \to Z$ (normed spaces) is continuous if and only if $\left\Vert B\left(x,y\right)\right\Vert \le C\left\Vert x\right\Vert \left\Vert y\right\Vert $ for all $x \in X, y \in Y$, where $C$ is some positive constant. The infimum over all such $C$ gives a norm on the vector space of bilinear transformation $X \times Y \to Z$. A particularly elegant way to see this is as follows:

A bilinear transformation $B: X \times Y \to Z$ is the same as a linear transformation $X\to\mathcal{L}\left(Y,Z\right)$. You can take the operator norm on $\mathcal{L}\left(Y,Z\right)$ and ask whether $x\mapsto\left[y\mapsto B\left(x,y\right)\right]$ is then a well-defined continuous transformation (note that for fixed $x \in X$, the map $y \mapsto B(x,y)$ is not a-priori continuous, i.e. in $\mathcal{L}\left(Y,Z\right)$). This is true if and only if $B$ is continuous and then the operator norm of this transformation will be the optimal $C$ aforementioned.

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$B(x,y)=xy$ is not a LINEAR operator from $\Bbb{R}^2$ to $\Bbb{R}$, although it is continuous. You can find easy counterexamples for this ($B(1,1), B(2,2)$).

The norm of a linear operator depends only the norm of the spaces where the operator is defined. If a continuous function is not bounded, then it surely is not linear, since for linear operators continuity and boundedness are equivalent concepts.

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