Margin bound for binary classification and Rademacher complexity In the slides (slide 29) of Mohri a margin bound for binary classification is derived:
$$R(h) \leq \hat{R}_\rho(h) + \frac{2}{\rho} \hat{R}_S(H) + 3\sqrt{\frac{\log \frac{2}{\delta}}{2m}}$$
Here $\hat{R}_\rho(h)$ is the empirical margin loss of $h$, $R(h)$ is the generalization error of $h$, and $\hat{R}_S(H)$ is the empirical Rademacher complexity of $H$. $\rho > 0$ is the desired margin. In all derivations of these kinds of bounds, it is mentioned that $\rho > 0 $ is fixed. 
1) I was wondering, can we compute this bound on the generalization error for multiple values of $\rho$, and then take the one that is the tightest? Or is this not allowed since $\rho$ is fixed?
2) Another question that I have is the following. Must we fix the hypothesis class $H$ before we can compute the empirical Rademacher complexity, or is it possible to specify the hypothesis class after seeing all samples? To be more precise, I want a hypothesis class of the following form:
$$H = \{x \mapsto <w,\Phi(x_i)>: w = f(K,x)\}$$
Where $f$ is a function that depends on the kernel matrix $K$ of all samples: $$K(i,j) = <\Phi(x_i),\Phi(x_j)>$$ and another vector $x$. To train the classifier, I obtain $x$ by optimizing a SRM problem:
$$h^* = \underset{h \in H}{\text{argmin}}~ \lambda ||h||^2 + \sum_i L(h(x_i))$$
Since $h$ only depends on $f(K,x)$ the above minimization can be rewritten as a minimization in terms of $x$. 
 A: 1) When you train your classifier or your regressor, the goal if to maximize $\rho$. It actually corresponds to the geometrical margin for the case of SVM classification!. So there is one value, the one you which is optimized.
Check slide 30. Here $\rho$, the margin, is introduced in an abstract way, using the properties of the Rademacher complexity to bound the empirical error, so you can have an estimate of how well your algorithm generalizes. Concretely, he introduces the Lipschitz function depicted in slide 28 (the hinge loss) to bind the 0-1 loss. It is just to get a bound on the Rademacher complexity.
Notice that, in principle, that value is arbitrary. But, in practice, you want the bound to be as tight as possible. And that corresponds to the maximum margin you are used to think of in geometrical terms.
What this bound is telling you is: given test data sampled from the same distribution as the training data (which is supposed to be stationary), how likely is that you perform as well on the new data as on the training data?. Is about analysing the stability of your classifier.
2) The hypothesis class is given by the choice you make of your kernel. Given the kernel, by the Mercer theorem, the $\Phi$'s are also determined.
