# gradient descent and local maximum

I read that gradient descent converge always to a local minimum while other methods as Newton's method this is not guaranteed (if the Hessian is not definite positive); but if the start point in GD is unfortunately a local maximum (and then the derivative is zero), how we can say that it converge to a minimum? does not it stall?

• As a comment: gradient descent is a minimization algorithm, so it searches for the minimum (or negative maximum if you need maximum). – Tim Aug 13 '18 at 12:48
• @Tim Your first comment seems to interpret "local minimum" as "global minimum." – whuber Aug 13 '18 at 13:05
• Can you please clarify this question? Your use of a concatenated "local/global" is confusing. Local and global minimum are very different and the answer to your question depends on what you mean. Most algorithms can trivially converge to a local minimum, but finding an algorithm that can efficiently converge to a global minimum is one of the biggest open questions in mathematics and theoretical CS. – Skander H. Aug 13 '18 at 18:52
• My question is just on the start point: if the start point is a maximum (so with gradient equals to zero) how can we say that the gradient descent go to a local minimum? – volperossa Aug 13 '18 at 18:54
• Providing a source for your claim would be helpful. I think that under certain conditions (decreasing step size, continuous function), GD is guaranteed to converge to a point with zero derivative (saddle point or extremum). If the initial point has already zero derivative (maximum), the algorithm has converged as expected. – Jan Kukacka Aug 14 '18 at 10:54