Lipschitz condition and differentiability
If and are Banach spaces, e.g. , one can inquire about the relation between differentiability and the Lipschitz condition. If is Lipschitz, the ratio
Proposition 1
Let be a continuously differentiable mapping (http://planetmath.org/DifferentiableMapping) between Banach spaces. If is a compact subset, then the restriction satisfies the Lipschitz condition.
Proof. Let denote the Banach space of bounded linear maps from to . Recall that the norm of a linear mapping is defined by
Let denote the derivative of . By definition is continuous, which really means that is a continuous function. Since is compact, there exists a finite upper bound for restricted to . In particular, this means that
for all .
Next, consider the secant mapping defined by
This mapping is continuous, because is assumed to be continuously differentiable. Hence, there is a finite upper bound for restricted to the compact set . It follows that for all we have
Therefore is the desired Lipschitz constant. QED
Neither condition is stronger. For example, the function given by is differentiable but not Lipschitz.
Title | Lipschitz condition and differentiability |
---|---|
Canonical name | LipschitzConditionAndDifferentiability |
Date of creation | 2013-03-22 11:57:50 |
Last modified on | 2013-03-22 11:57:50 |
Owner | Mathprof (13753) |
Last modified by | Mathprof (13753) |
Numerical id | 34 |
Author | Mathprof (13753) |
Entry type | Theorem |
Classification | msc 26A16 |
Synonym | mean value inequality |
Related topic | Derivative2 |