Since $(x-a)^2 \geq 0$ you can rewrite $h$ to $h(x) = \max\{(x-a)^2, (1-ax)(x-a)^2\}$. Since the second function is only larger than the first one if $ax \leq 0$, you can replace its value for $ax\geq 0$ without changing $h$. Define

$$g(x) = \begin{cases}(1-ax)(x-a)^2 & \text{if } ax \leq 0 \\ a^2 - (a^3+2a)x & \text{else.} \end{cases}$$ Since the first branch is convex, and the second branch is just the tangent of the first branch at $x=0$, the function $g$ is convex. Now $h(x) = \max\{(x-a)^2, g(x)\}$, i.e., the maximum of two convex functions, and therefore convex.

Here's my attempt to answer your questions:

• Is an SVM as simple as saying it's a discriminative classifier that simply optimizes the hinge loss? Or is it more complex than that? Yes, you can say that. Also, don't forget that it regularizes the model too. I wouldn't say SVM is more complex than that, however, it is important to mention that all of those choices (e.g. hinge loss and $L_2$ regularization) have precise mathematical interpretations and are not arbitrary. That's what makes SVMs so popular and powerful. For example, hinge loss is a continuous and convex upper bound to the task loss which, for binary classification problems, is the $0/1$ loss. Note that $0/1$ loss is non-convex and discontinuous. Convexity of hinge loss makes the entire training objective of SVM convex. The fact that it is an upper bound to the task loss guarantees that the minimizer of the bound won't have a bad value on the task loss. $L_2$ regularization can be geometrically interpreted as the size of the margin.

• How do the support vectors come into play? Support vectors play an important role in training SVMs. They identify the separating hyperplane. Let $D$ denote a training set and $SV(D) \subseteq D$ be the set of support vectors that you get by training an SVM on $D$ (assume all hyperparameters are fixed a priori). If we throw out all the non-SV samples from $D$ and train another SVM (with the same hyperparameter values) on the remaining samples (i.e. on $SV(D)$) we get the same exact classifier as before!

• What about the slack variables? SVM was originally designed for problems where there exists a separating hyperplane (i.e. a hyperplane that perfectly separates the training samples from the two classes), and the goal was to find, among all separating hyperplanes, the hyperplane with the largest margin. The margin, denoted by $d(w, D)$, is defined for a classifier $w$ and a training set $D$. Assuming $w$ perfectly separates all the examples in $D$, we have $d(w, D) = \min_{(x, y) \in D} y \frac{w^Tx}{||w||_2}$, which is the distance of the closest training example from the separating hyperplane $w$. Note that $y \in \{+1, -1\}$ here. The introduction of slack variables made it possible to train SVMs on problems where either 1) a separating hyperplane does not exist (i.e. the training data is not linearly separable), or 2) you are happy to (or would like to) sacrifice making some error (higher bias) for better generalization (lower variance). However, this comes at the price of breaking some of the concrete mathematical and geometric interpretations of SVMs without slack variables (e.g. the geometrical interpretation of the margin).

• Why can't you have deep SVM's? SVM objective is convex. More precisely, it is piecewise quadratic; that is because the $L_2$ regularizer is quadratic and the hinge loss is piecewise linear. The training objectives in deep hierarchical models, however, are much more complex. In particular, they are not convex. Of course, one can design a hierarchical discriminative model with hinge loss and $L_2$ regularization etc., but, it wouldn't be called an SVM. In fact, the hinge loss is commonly used in DNNs (Deep Neural Networks) for classification problems.