# Proving $\|f(x)-f(a)\|\le M\| x-a\|$

Theorem(1): (M.V.T for real valued functions)

Let $V \subseteq \Bbb R^n$ be open. Suppose $f: V \to \Bbb R$ is differentiable on $V$. If $x, a\in V$ then there is $c\in L(x;a)$ such that $f(x)-f(a)=\nabla f(c)\cdot (x-a)$

Theorem(2): (M.V.T for vector valued functions)

Let $V \subseteq \Bbb R^n$ be open. Suppose $f: V \to \Bbb R^m$ is differentiable on $V$. If $x, a\in V$ and $L(x;a)\subseteq V$ then given any u $\in \Bbb R^m$there is $c\in L(x;a)$ such that (u)$(f(x)-f(a))=$ (u) $(Df(c)\cdot (x-a))$

The statement:

Let $V \subseteq \Bbb R^n$ be open. Let $H$ be compact subset of $V$ and suppose that $f: V \to \Bbb R^m$ is $C^1$ on $V$. If $E\subset H$ is convex, then there is a constant $M$ such that $\| f(x)-f(a)\|\le M\| x-a\| \forall x,a \in E.$

I need to prove that the statement by using the theorem(1) -not using the theorem (2)-

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There probably are two typos in the last inequality. –  Git Gud Jul 16 '13 at 19:12
And in the title also. ($x \mapsto a$) –  Cortizol Jul 16 '13 at 19:13

Hint: apply theorem 1 to each component of $f=(f_1,\dots,f_m)$.

Edit: elaborating

Clearly, each component $f_i:V\to\mathbb R$ is $\mathcal C^1$ on $V$. Thus, we can apply th. 1 to say that $\exists c_i\in L(a,x)$ (note that $c_i\in E$, thanks to convexity), such that $f_i(x)-f_i(a)=\nabla f_i (c)\cdot(x-a)$. This leads to say that $$|f_i(x)-f_i(a)|\le \|\nabla f_i (c)\|\|x-a\|.$$ As $\nabla f_i$ is a continuous function on a compact, there's a supremum of $\|\nabla f_i\|$. Let $$M_i=\sup_{c\in E}\|\nabla f_i(c)\|,$$ then $$\forall x,a \in E\quad |f_i(x)-f_i(a)|\le M_i\|x-a\|.$$ Then again, let $$M=\frac{1}{\sqrt m}\max_{i=1..n} M_i, so$$ $$\|f(x)-f(a)\|^2 = \sum_i |f_i(x)-f_i(a)|^2\le \frac{M^2}{m} \sum_i \| x - a \|^2 =M^2\| x - a \|^2.$$

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I would want to do, but I could not do. Please can you show step by step. Thank you so much. –  Mathlover4 Jul 16 '13 at 19:13
(+1) Just note: We proved this for Euclidean norm, but that is not problem because all norms on finite-dimensional vector spaces are equivalent. –  Cortizol Jul 16 '13 at 19:36
Yeah! I understand well. Thank you:)) –  Mathlover4 Jul 16 '13 at 21:48

Let $f=(f_1,...f_m).$Since $E$ is convex, for any $x,y\in E$, $L(x;y) \subset E \subset K .$ As your given $f$ is $C^1$ on $V$, Note that $\nabla f_i$ is continuous on $V$ for each $i$.Choose $M'=max_{1\leq i\leq m}[ {\sup_{x\in K}||\nabla f_i(x)||}]$.As $K$ is compact subset of $V$, $M'$ will be finite.And for this $M'$ clearly $||f(x)-f(y)||\space = \space (\sum _{i}||f_i(x)-f_i(y)||^2)^{1/2}\space \leq \space [\sum _{i}||\nabla f_i(c_{x,y})||^2]^{1/2}.||x-y|| \space \space$ $\leq \space m^{1/2}M'||x-y|| \forall x,y \in E.$

Now chose your correct $M$.

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