# Lipschitz condition and differentiability

If $X$ and $Y$ are Banach spaces^{}, e.g. ${\mathbb{R}}^{n}$, one can inquire about the relation^{}
between differentiability and the Lipschitz condition^{}. If $f$ is Lipschitz, the ratio

$$\frac{\parallel f(q)-f(p)\parallel}{\parallel q-p\parallel},p,q\in X$$ |

###### Proposition 1

Let $f\mathrm{:}X\mathrm{\to}Y$ be a continuously differentiable mapping (http://planetmath.org/DifferentiableMapping) between
Banach spaces. If $K\mathrm{\subset}X$ is a compact
subset, then the restriction^{} $f\mathrm{:}K\mathrm{\to}Y$ satisfies the Lipschitz
condition.

*Proof.*
Let $\mathrm{lin}(X,Y)$ denote the Banach space of bounded linear maps from
$X$ to $Y$. Recall that the norm $\parallel T\parallel $ of a linear mapping
$T\in \mathrm{lin}(X,Y)$ is defined by

$$\parallel T\parallel =sup\{\frac{\parallel Tu\parallel}{\parallel u\parallel}:u\ne 0\}.$$ |

Let $\mathrm{D}f:X\to \mathrm{lin}(X,Y)$ denote the derivative^{} of $f$. By definition
$\mathrm{D}f$ is continuous^{}, which really means that
$\parallel \mathrm{D}f\parallel :X\to \mathbb{R}$
is a continuous function. Since
$K\subset X$ is compact, there exists a finite upper bound ${B}_{1}>0$ for
$\parallel \mathrm{D}f\parallel $ restricted to $K$. In particular, this means that

$$\parallel \mathrm{D}f(p)u\parallel \le \parallel \mathrm{D}f(p)\parallel \parallel u\parallel \le {B}_{1}\parallel u\parallel ,$$ |

for all $p\in K,u\in X$.

Next, consider the secant mapping $s:X\times X\to \mathbb{R}$ defined by

$$s(p,q)=\{\begin{array}{cc}\frac{\parallel f(q)-f(p)-\mathrm{D}f(p)(q-p)\parallel}{\parallel q-p\parallel}\hfill & q\ne p\hfill \\ 0\hfill & p=q\hfill \end{array}$$ |

This mapping is continuous, because $f$ is assumed to be continuously differentiable. Hence, there is a finite upper bound ${B}_{2}>0$ for $s$ restricted to the compact set $K\times K$. It follows that for all $p,q\in K$ we have

$\parallel f(q)-f(p)\parallel $ | $\le \parallel f(q)-f(p)-\mathrm{D}f(p)(q-p)\parallel +\parallel \mathrm{D}f(p)(q-p)\parallel $ | ||

$\le {B}_{2}\parallel q-p\parallel +{B}_{1}\parallel q-p\parallel $ | |||

$=({B}_{1}+{B}_{2})\parallel q-p\parallel $ |

Therefore ${B}_{1}+{B}_{2}$ is the desired Lipschitz constant. QED

Neither condition is stronger. For example, the function $f:\mathbb{R}\to \mathbb{R}$
given by $f(x)={x}^{2}$ 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 |