In the context of kernel methods, this is quite easy to see. Any given data matrix $\mathbf{X}$ corresponds to a kernel matrix $\mathbf{K}$, which in turns corresponds to a certain solution of the training problem which, due to convexity, is unique and guaranteed to be the global optimum.
However, if we change $\mathbf{X}$ into $\mathbf{X}'$, via some transformation, then we get a different kernel matrix $\mathbf{K}'$ and evidently a different solution. I explicitly mention this, because it is important to realize that the solution changes when we change $\mathbf{X}$, for instance by scaling. The solution is still found by solving a convex problem, and hence is still unique and the global optimum of its corresponding training problem (which is different for $\mathbf{K}$ and $\mathbf{K}'$).
The reason we do scaling, then, has nothing to do with solving the optimization problem but rather with defining it. For simplicity, lets assume we use the standard linear kernel: $$\kappa(\mathbf{u},\mathbf{v}) = \mathbf{u}^T\mathbf{v}$$ If the features are on different scales, say, the first dimension is in $[0, 10^{99}]$ and the second is in $[0, 1]$, then it is easy to see that in all these kernel evaluations the first feature will completely dominate in the resulting distance estimates (that is, entries in the kernel matrix are almost exclusively based on the first dimension).
When you start building a model, you typically want to give each feature a similar contribution in the model, and for kernel methods that implies that they should be on the same scale. Scaling in no way guarantees better performing models, but it is usually the best "prior" you have. For example assume that the first feature in the contrived example above is informative while the second is not, in this case leaving the first feature on a far larger scale is in fact better.
