# Meaning of product-like construction in Hilbert space

In [1, Assumption 2] the expression $$|\nabla^3 \ell_z(\theta)[k,h,h]|\leq \sup_{g\in\mathcal{H}}|k\cdot g| \nabla^2\ell_z(\theta)[h,h]$$ is used, where $$\ell_z:\mathcal{H}\to \mathbb{R}$$ with a reproducing kernel Hilbert space $$\mathcal{H}$$.

I don't understand the meaning of $$[k,h,h]$$ and cannot find a definition in the paper. Do I miss some standard construction in Hilbert spaces?

Here, $$\ell$$ (I drop the suffix $$z$$ for convenience) is defined on some Hilbert space $$\mathcal H$$ which has no reason to be $$\mathbb R^d$$. Hence its gradient $$\nabla\ell$$, Hessian $$\nabla ^2\ell$$, and all higher order derivatives $$\nabla ^n\ell$$ (assuming they exist) need to be defined a little bit differently than in the Euclidean case.
• Let's start with the gradient $$\nabla \ell$$ : the direct analogue of the gradient for functions defined on Hilbert spaces would be the Fréchet derivative. For $$\ell : \mathcal H \to \mathbb R$$ and $$\theta\in\mathcal H$$, we define the Fréchet derivative of $$\ell$$ at $$\theta$$ as the unique bounded linear functional $$\nabla \ell(\theta) : \mathcal H \to \mathbb R$$ such that, $$\lim_{h\in\mathcal H,\|h\|\to 0}\frac{\|\ell(\theta+h) - \ell(\theta) - \nabla\ell(\theta)[h]\|}{\|h\|} = 0$$ Hence you see that for any point $$\theta\in\mathcal H$$, $$\nabla\ell(\theta)$$ is a linear functional defined on $$\mathcal H$$ rather than a vector. However, Riesz representation theorem tells us that for any bounded linear functional $$f$$ on $$\mathcal H$$, there exists a unique $$f^*\in\mathcal H$$ such that $$f(x) = \langle f^*,x\rangle$$ for all $$x\in \mathcal H$$. In the special case $$\mathcal H = \mathbb R^d$$, this tells us that the Fréchet derivative $$Df(x_0)$$ of a differentiable map $$f:\mathbb R^d \to \mathbb R$$ at a point $$x_0\in \mathbb R^d$$ is given by $$Df(x_0) : x \mapsto \langle \nabla f(x_0),x \rangle$$, where $$\nabla f(x_0)$$ is the usual gradient of $$f$$ at $$x_0$$. All that is to say that, indeed, the Fréchet derivative extends the definition of gradient to general Hilbert spaces (normed spaces, actually).
• If you get the first definition, then it is not hard to extend it to define the Hessian $$\nabla^2\ell$$ : For some parameter $$\theta\in\mathcal H$$ and $$\ell :\mathcal H \to \mathbb R$$, we define the Hessian of $$\ell$$ at $$\theta$$ as the Fréchet derivative of $$\nabla \ell (\theta)$$, i.e. the unique bounded linear operator $$\nabla ^2\ell(\theta) : \mathcal H \to \mathscr L(\mathcal H,\mathbb R)$$ such that $$\lim_{h\in\mathcal H,\|h\|\to 0}\frac{\|\nabla\ell(\theta+h) - \nabla\ell(\theta) - \nabla^2\ell(\theta)[h]\|}{\|h\|} = 0$$ Notice in particular that $$\nabla^2\ell(\theta)[h]$$ is not a real number but rather a bounded linear functional defined on $$\mathcal H$$. This implies by Riesz theorem that given a fixed $$\theta \in \mathcal H$$, for all $$h\in\mathcal H$$, there exists a unique element $$\text{Hess}(\theta)[h]\in\mathcal H$$ such that $$\nabla ^2\ell(\theta)[h][x] \equiv \nabla ^2\ell(\theta)[h,x] = \langle \text{Hess}(\theta)[h], x \rangle$$ This implies that $$\nabla^2\ell(\theta)$$ can be uniquely identified as a bilinear map defined on $$\mathcal H \times \mathcal H$$
• If you understand the above then it's not hard to see (but tedious to write) how to generalize the above definitions to higher order derivatives of $$\ell$$, and you should be able to convince yourself that for any integer $$n\ge 1$$, $$\nabla^n\ell$$ is a real-valued multilinear map defined on the product space $$\mathcal H\times \ldots \times \mathcal H$$.
That was a lot, but if you followed you should now be able to answer your question : In the equation you write, the terms in bracket $$[k,h,h]$$ and $$[h,h]$$ do not represent a "product", but they are rather the respective arguments of the multilinear maps $$\nabla^3\ell_z(\theta)$$ and $$\nabla^2\ell_z(\theta)$$, defined on $$\mathcal H \times \mathcal H \times\mathcal H$$ and $$\mathcal H \times\mathcal H$$ respectively.