# $C^1$ function on compact set is Lipschitz

Let $\emptyset\ne A\subset\mathbb{R}^n$ be open and let $f \in C^1 (A, \mathbb{R})$ be a function. Let $\emptyset\ne K\subset A$ be compact and convex. I want to prove that $f$ is Lipschitz on $K$; that is, prove there exists a constant $c > 0$ such that $| f ( x ) - f( y ) | \leq c \, \| x - y \|, \forall x , y \in K$.

My approach:

Let $x,y\in K$ be two arbitrary points. Then, since $K$ is convex, the line segment between $x$ and $y$, i.e. $Co(x,y)$, is in $K$. Thus, by MVT, there exists a vector $b\in Co(x,y)$ such that

$$\text{(*) } |f(x)-f(y)|=|\langle x-y, (\nabla f)(b)\rangle|\le \|x-y\|\|(\nabla f)(b)\|$$

Now, since $K$ is compact, $f$ takes a maximum and a minimum values on $K$, so that $\exists b'\in K$ such that $\|(\nabla f)(b') \|\ge \|(\nabla f)(b) \|$, for all $b\in K$. Let $c\in \mathbb{R}$, $c:=(\nabla f)(b')$, then

$$|f(x)-f(y)|\le\|x-y\|\|(\nabla f)(b)\|\le\|x-y\|\|(\nabla f)(b')\|=c\|x-y\|$$

This implies that $f$ is Lipschitz on $K$ for all $x,y\in K$.

Please let me know if you think my proof is correct or not very much? I'm somewhat concerned about the part with the gradient - how exactly is the maximality of the norm of the gradient related to the EVT, that is to $f$ taking maximum and minimum values? As far as I can tell, the maximum norm of the gradient exists because that is the direction to the maximum (or minimum) point of $f$.

• What is $a$? It seems undefined to me . . . – Robert Lewis Jun 28 '17 at 0:24
• Sorry, it was supposed to be $y$. Fixed. – sequence Jun 28 '17 at 0:26
• I guess is because $||f'||$ is a real function on a compact set? – Li Chun Min Jun 28 '17 at 0:41
• @LiChunMin Can you please clarify your question? – sequence Jun 28 '17 at 0:47
• Looks good to me. You do not need the EVT. Since $\nabla (f)$ is continuous, the real-valued function $\|\nabla (f)\|$ is continuous .So $T=\{\| \nabla f(x)\|: x\in K\}$ is compact (because $K$ is compact) , so $T$ is a bounded real subset....So $\|f(x)-f(y)\|\leq \|x-y\|\cdot \sup T$ for all $x,y\in K$. – DanielWainfleet Jun 28 '17 at 2:48

Here's how I would do it:

for $x, y \in K$, let $\gamma(t):[0, 1] \to K$ be the line segment

$\gamma(t) = x + t(y - x); \tag{1}$

then

$\gamma(0) = x, \tag{2}$

$\gamma(1) = x +(y - x) = y, \tag{3}$

and

$\gamma'(t) = y - x; \tag{4}$

furthermore, since $K$ is convex, $\gamma(t)$ lies entirely within $K$, hence also in $A$.

Now, for $x, y \in K$, we have:

$\Vert f(y) - f(x) \Vert = \Vert \displaystyle \int_0^1 \dfrac{d(f(\gamma(t))}{dt}dt \Vert = \Vert \displaystyle \int_0^1 \nabla f(\gamma(t)) \cdot \gamma'(t) dt \Vert$ $\le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \cdot \gamma'(t) \Vert dt \le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt.\tag{5}$

Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$, hence

$\displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt \le \displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt; \tag{6}$

using (4),

$\displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt = \displaystyle \int_0^1 B \Vert y - x \Vert dt = B \Vert y - x \Vert; \tag{7}$

bringing together (5), (6), and (7) yields

$\Vert f(y) - f(x) \Vert \le B \Vert y - x \Vert, \tag{8}$

that is, $f(x)$ is Lipschitz continuous on $K$.

• Thank you, this is very elegant. Do you think my proof could possibly be correct? – sequence Jun 28 '17 at 1:32
• Yes, your method seems fine. Note it agrees with mine. – Robert Lewis Jun 28 '17 at 1:39
• "Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$.." Is there a proof for that or is it trivial? – abuchay May 14 '18 at 14:25
• @abuchay: Well, $\nabla f \in C^0(K)$ since $\nabla f \in C^0(A)$ and $K \subset A$. $\Vert \nabla f \Vert \in C^0(K)$ since $\Vert \cdot \Vert$ is itself continuous in its argument. Then the boundedness follows since continuous functions on compacta are bounded, which is standard result in elementary topology. – Robert Lewis May 14 '18 at 15:40

The convexity of $K$ is not needed. Suppose the conclusion fails. Then for each $m\in \mathbb N,$ there exist $y_m,x_m \in K$ such that

$$\tag 1 |f(y_m)- f(x_m)| > m|y_m-x_m|.$$

Because $K$ is compact, we can find a subsequence $m_k$ such that the sequences $y_{m_k},x_{m_k}$ converge to points $y,x\in K$ respectively.

Suppose $y\ne x.$ Then $|y-x| > 0.$ For large $k$ we then have

$$|f(y_{m_k})- f(x_{m_k})| > m_k|y_{m_k}-x_{m_k}|> m_k(|y-x|/2)\to \infty$$

This implies $f$ is not bounded on $K.$ But $f$ is continuous on $A,$ hence is continuous on $K,$ hence $f$ is bounded on $K$ by compactness. This contradiction shows $y=x.$

Because $A$ is open we can choose $r>0$ such that $B(x,r)\subset A.$ For large $k$ we then have $y_{m_k},x_{m_k}$ in the compact convex set $\overline {B(x,r/2}) \subset A.$ By $(1),$ your highlighted mean value inequality then fails in this last set, contradiction. Therefore $(1)$ cannot hold, proving the result.

• I dont understand why $m_k(|y-x|/2)\to \infty$. Why does it tend to infinity? If $m_k$ is a subsequence of $y_{mk}$ then $m_k$ also converges to $y$. Then, how can $m_k$ be a subsequence of $x_{mk}$ too? – John Keeper Oct 24 '18 at 9:46
• @JohnKeeper $m_k$ is a subsequence of $1,2,\dots$ hence $\to \infty$ – zhw. Oct 24 '18 at 21:32