Why setting SVDD's C-parameter to $> 1$ does not affect the result? Why setting $C>1$ does not affect the result (compared to $C=1$) according to:
https://www.csie.ntu.edu.tw/~cjlin/libsvmtools/#libsvm_for_svdd_and_finding_the_smallest_sphere_containing_all_data
However in their reference paper [1] the optimization is 
$$\min_{R,\mathbf{a},\boldsymbol\eta}\ (R^2 + C \sum_{i=1 \dots \ell} \eta_i)$$
subject to
$$(\phi(\mathbf{x}_i)-\mathbf{a})^2 \leq R^2 + \eta_i,\ \ \ \eta_i> 0, \ \ \ i = 1,\dots,\ell $$.
[1] https://www.csie.ntu.edu.tw/~cjlin/papers/svdd.pdf
   A Revisit to Support Vector Data Description
 A: So this is Theorem 6 in that paper. They then provide the following text:
For SVM, it is known that if the training instances are separable, there is a $\bar{C}$ such that for all $C > \bar{C}$, the solution is the same
as that of the problem without the loss term; see, for example, Lin (2001). This $\bar{C}$ is
problem dependent, but for SVDD, we have shown that $\bar{C}$ is one.
The proof of this basically relies on the fact that when $C \geq \frac{1}{\ell}$, that $0 \leq \alpha_i \leq 1$ (because $\sum \alpha_i = 1$ in this case). This in turn implies that $\gamma_i > 0$ (since $C = \alpha_i + \gamma_i$) which then results in $\xi_i = 0$, in other words the data is "separable".
Since this is for SVDD - this is saying that basically for $C > 1$, the result is always the same - namely the smallest sphere containing all of the data. 
Note though that this is for the version of SVDD that has $\bar{R}$ and not $R^2$ in terms of the optimization (they explain numerous issues with the version with $R^2$).
To give some additional intuition into the result, consider that $\sum_i \alpha_i = 1$ at the optimum (for $C \geq \frac{1}{\ell}$). Then consider that for all points which lie outside the sphere, $\xi_i > 0 \Rightarrow \gamma_i = 0 \Rightarrow \alpha_i = C$. Therefore, the number of points that can lie outside the sphere is at most $\frac{1}{C}$. Thus, if $C > 1$, the number of points that can lie outside the sphere must be less than 1. Thus, when $C > 1$, no points can lie outside the sphere, and thus the solution is always the smallest sphere(s) containing all of the points (regardless of uniqueness of the result).
