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?