Understanding Extremum and Derivatives at Boundary Points

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Leo Authersh
It's understandable that finding absolute extremum is impossible for a function with restricted boundary conditions. But why does the derivative of similar functions is not zero when the extremum is on the end points?

To be precisely short with my question, why does the derivative gives only the extremum at the interior points within the boundary and not at the points on the boundary?
 
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In short, assuming that the function is continuous, because it does not matter what the derivative is on the boundary. Regardless of the derivative the value of the function at the boundary will be larger than or smaller all other function values in a small region close to the boundary.

Note that this holds only for one-dimensional functions. If you deal with functions in several dimensions and the boundary has a dimension greater than zero, then the derivative along the boundary must be zero for there to be an extremum on the boundary.
 
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Orodruin said:
In short, assuming that the function is continuous, because it does not matter what the derivative is on the boundary. Regardless of the derivative the value of the function at the boundary will be larger than or smaller all other function values in a small region close to the boundary.

Note that this holds only for one-dimensional functions. If you deal with functions in several dimensions and the boundary has a dimension greater than zero, then the derivative along the boundary must be zero for there to be an extremum on the boundary.
This is the topological boundary, right? Aren't Topological boundaries necessarily zero-dimensional? Or do you mean Manifold boundaries?
 
Why would the topological boundary need to be zero dimensional? Consider the topological boundary of the set ##x^2+y^2\leq 1##, which is a circle. Anyway, an important issue is that the boundary point must also be in the set in order to provide a local extremum.

However, I believe this discussion may be too advanced for a B thread.
 
Orodruin said:
Why would the topological boundary need to be zero dimensional? Consider the topological boundary of the set ##x^2+y^2\leq 1##, which is a circle. Anyway, an important issue is that the boundary point must also be in the set in order to provide a local extremum.

However, I believe this discussion may be too advanced for a B thread.
Sorry, you're right, I think it was some measure of " meagreness" in the ambient space. But you're right, let's drop it and save it for some other post.