I am confronted with following problem: Let $\beta\in \Bbb R$ and $y_k\in\Bbb R^n$ for every $k$.
minimise $\alpha+\frac{1}{q(1-\beta)}\sum_{k=1}^{q}max(-x^Ty_k-\alpha,0)$
subject to: $x_j\geq0$ for $j=1,...,n$ with $\sum_{j=1}^nx_j=1$ and $\alpha\in\Bbb R$
minimise $\alpha+\frac{1}{q(1-\beta)}\sum_{k=1}^{q}u_k$
subject to: $x_j\geq0$ for $j=1,...,q$ with $\sum_{j=1}^nx_j=1$, $\alpha\in\Bbb R, u_k\geq0$ and $x^Ty_k+\alpha+u_k\geq 0$
This both problem should be equivalent. However I failed to recognise this as in problem (2) the last condition can be rewritten as $u_k\geq-x^Ty_k-\alpha$. Combining with $u_k\geq0$ this would mean that I am allowing $-x^Ty_k-a\leq0$ in the sum, which is 0 in the problem (1).
This is called the "epigraph form" of an optimization problem. For any $\alpha > 0$, the problem $$ \begin{aligned} \min_x &\quad f(x) + \alpha g(x) \\ \text{s.t.} &\quad x \in C \end{aligned} $$ can be equivalently written as $$ \begin{aligned} \min_{x, u} &\quad f(x) + \alpha u \\ \text{s.t.} &\quad g(x) \leq u \\ &\quad x \in C \end{aligned} $$ The equivalence is quite easy to establish simply by following the definition of the optimum. Thus, any term in the objective function can be "moved to the constraints". Applying this rule to every one of the terms $\max(-x^T y_k - \alpha, 0)$ in problem (1) you wrote, together with the fact that $$ \max(a, b) \leq c \iff a \leq c \land b \leq c $$ yields problem (2).