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 |