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I am given the following expression where $f \in L^2(\mathbb{R}^2, \mathbb{R}^{2 \times 2})$

$$\int_{\mathbb{R}} \int_{\mathbb{R}} \langle g(x), f(x,y) h(y)\rangle dx dy.$$

The functions $g$ and $h$ are in $L^2(\mathbb{R},\mathbb{R}^2).$

Question: Is it then true that

$$\sup_{\left\lVert g \right\rVert=\left\lVert h \right\rVert=1}\int_{\mathbb{R}} \int_{\mathbb{R}} \langle g(x), f(x,y) h(y)\rangle dx dy = \left\lVert f \right\rVert_{L^2(\mathbb{R}, \mathbb{R}^{2 \times 2})}^2?$$

Here $L^2(\mathbb{R}, \mathbb{R}^{2 \times 2})$ is the $L^2$ norm of the operator norm of $f.$

Background: I would like to explain the question by saying that it is true that

$$\langle v, Aw\rangle$$ maximized over all $v,w$ of norm $1$ for a matrix $A$ gives the operator norm of $A$ and calculating

$$\int_{\mathbb{R}} \int_{\mathbb{R}} \overline{h(x,y)} f(x,y) dx dy.$$

for $f \in L^2(\mathbb{R}^2; \mathbb{R})$ maximized over all $h \in L^2(\mathbb{R}^2; \mathbb{R})$ of norm one yields the $L^2$ norm of $f$.

But is this mixed result above true as well?

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The inequality $\leq$ is obvious.

The reverse doesn't have to be true.

Let $f(x,y) = \mathrm{Id} \cdot \chi(x,y)$ where $\chi(x,y)$ is the scalar function that equals $1$ on the squares $$ (x,y) \in [0,1] \times [0,1] \cup [-1,0]\times [-1,0] $$ and zero otherwise.

So $$ \iint_{\mathbb{R}\times\mathbb{R}} \langle g(x), f(x,y) h(y) \rangle ~\mathrm{d}x ~\mathrm{d}y = \int_0^1 \int_0^1 \langle g(x), h(y) \rangle ~\mathrm{d}x ~\mathrm{d}y \\ + \int_{-1}^0 \int_{-1}^0 \langle g(x), h(y) \rangle ~\mathrm{d}x ~\mathrm{d}y$$ Using Holder's inequality we can bound the right hand side by $$ \leq G_+ H_+ + G_- H_- $$ where $$ G_+ = \|g\|_{L^2([0,1])}, H_+ = \|h\|_{L^2([0,1])}$$ $$ G_- = \|g\|_{L^2([-1,0])}, H_- = \|h\|_{L^2([-1,0])} $$ By assumption $G_+^2 + G_-^2 \leq 1$ and also $H_+^2 + H_-^2 \leq 1$. So by Cauchy's inequality we have that $$ G_+ H_+ + G_- H_- \leq 1$$

On the other hand, the operator norm of $f(x,y)$, as a function of $x,y$, is simply the function $\chi(x,y)$. This function has $L^2(\mathbb{R}^2)$ norm equal to $\sqrt{2} > 1$.


Remark: the problem is not so much with "integration" per se. As you can guess from the example given, the problem also manifests in the finite dimensional case, where $g$ and $h$ are functions $\{0,1\} \to \mathbb{R}^2$ and $f(x,y)$ is a function from $\{(0,0), (0,1), (1,0), (1,1)\} \to \mathbb{R}^{2\times 2}$, and the integral is replaced by a sum.

Thinking about this more, you will see that essentially we are taking a tensor product, and then you see that the problem is precisely that "not every element in $V\otimes W$ can be written as the tensor product of elements $v\otimes w$. (It can be written as a linear combination of such elements.)

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  • $\begingroup$ thank you very much for this insight. would it hold then if $f(x,y) = f_1(x)f_2(y)$ or do you see any (natural) condition on $f$ for this to be true? $\endgroup$ – user121558 Mar 8 '18 at 18:24
  • $\begingroup$ It is certainly true (rather trivially) if the matrix $f(x,y)$ splits as a tensor product $f_1(x) \otimes f_2(y)$, because then $ \langle g(x), f(x,y) h(y)\rangle = \langle g(x), f_1(x)\rangle \langle f_2(y) ,h(y)\rangle$. It is also true of $f(x,y) = M \cdot \chi(x) \cdot \eta(y)$ where $\chi, \eta$ are scalar functions and $M$ is a fixed $2\times 2$ matrix. For other notions of products you have to check it yourself. $\endgroup$ – Willie Wong Mar 8 '18 at 18:33
  • $\begingroup$ thanks, it also seems to hold if $f \in L^2(\mathbb{R}^2;\mathbb{R}^2)$,$g \in L^2(\mathbb{R},\mathbb{R}^2)$ and $h \in L^2(\mathbb{R},\mathbb{R}),$ right? So when we have a vector rather than a matrix, no? $\endgroup$ – user121558 Mar 8 '18 at 21:48
  • $\begingroup$ how do you multiple vectors? $\endgroup$ – Willie Wong Mar 9 '18 at 1:07
  • $\begingroup$ sorry, $f$ is a vector, $h$ a scalar and $g$ a vector-valued function. So the inner-product between the vector-valued functions $fh$ and $g$ is defined, no? $\endgroup$ – user121558 Mar 9 '18 at 1:42

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