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# 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

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

###### Proposition 1.

Let $f:X\to Y$ be a continuously differentiable mapping between Banach spaces. If $K\subset X$ is a compact subset, then the restriction $f:K\to Y$ satisfies the Lipschitz condition.

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

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

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

$\|{\operatorname{D}f}(p)u\|\leq\|{\operatorname{D}f}(p)\|\|u\|\leq B_{1}\|u\|,$ |

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

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

$s(p,q)=\begin{cases}\displaystyle\frac{\|f(q)-f(p)-{\operatorname{D}f}(p)(q-p)% \|}{\|q-p\|}&q\neq p\\ 0&p=q\end{cases}$ |

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

$\displaystyle\|f(q)-f(p)\|$ | $\displaystyle\leq\|f(q)-f(p)-{\operatorname{D}f}(p)(q-p)\|+\|{\operatorname{D}% f}(p)(q-p)\|$ | ||

$\displaystyle\leq B_{2}\|q-p\|+B_{1}\|q-p\|$ | |||

$\displaystyle=(B_{1}+B_{2})\|q-p\|$ |

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.

## Mathematics Subject Classification

26A16*no label found*

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## Attached Articles

## Corrections

U -> K by mathforever ✓

Lipschitz constant by CWoo ✓

x^2 by paolini ✓

proof supplied by Wkbj79 ✓

## Comments

## Proof contained

Since this entry contains its own proofs, it should be marked as such.

## Re: Proof contained

shouldn't this be filed as a "meta" correction to the entry?

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: Proof contained

Yes. My mistake.