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Prove that if $f$ is differentiable on $[a,b]$ and $f$ is Lipschitz continuous (LC), then $f$ has a bounded derivative.

My proof:

$f$ is LC $\Rightarrow$ f has bounded a derivative: there exists $M\gt 0$ such that $$|f(x)-f(y)|\le M|x-y|, \forall x,y\in [a,b]$$

then $${|f(x)-f(y)|\over {|x-y|}}\le M$$(if $ x\neq y$)

If $y\lt x$ (without loss of generality $x\lt y$) then $[y,x]\subseteq [a,b]$ and by hypothesis $f$ is differentiable on $[a,b]$ hence $f$ is differentiable on $[y,x]$. Then, using Mean Value Theorem there must exist $c\in (y,x)$ such that $$f´(c)={f(x)-f(y)\over x-y}$$ hence $$|f´(c)|\le M$$

As $x,y$ are arbitrary elements of $[a,b]$, then $c$ is also an arbitrary element but of $(a,b)$ therefore $$|f´(c)|\le M$$ $$\forall c\in (a,b)$$

but does this imply that $$|f´(c)|\le M, \forall c\in [a,b]?$$

I would really appreciate your help

mxdg
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user128422
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  • Do you mean for this to be a bidirectional (if and only if) statement? Your title reads weird otherwise. – Cameron Williams Mar 23 '14 at 21:06
  • I have already edited the title is not if only if, is just then – user128422 Mar 23 '14 at 21:07
  • but my question is that if $|f´(c)|\le M$ $\forall c\in (a,b)$ implies that $|f´(c)|\le M$ $\forall c\in [a,b]$ – user128422 Mar 23 '14 at 21:09
  • Do you want to say that if $f$ is differentiable on $[a,b]$ then it is Lipschitz AND has bounded derivative? Or: if $f$ is differentiable and Lipschitz then it has bounded derivative? I can't quite follow the work since the statement isn't clear to me. – Cameron Williams Mar 23 '14 at 21:10
  • is the second one: If $f$ is differentiable and lipschitz then it has bounded derivative – user128422 Mar 23 '14 at 21:12

2 Answers2

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Your "proof" could be used for the converse: When $|f'(x)|\leq M$ for all $x\in[a,b]$ then $f$ is Lipschitz continuous on $[a,b]$, and $M$ can serve as a Lipschitz constant.

In fact we are asked to prove that there is an $M>0$ with $|f'(x)|\leq M$ for all $x\in[a,b]$, given that $f$ is Lipschitz continuous on $[a,b]$.

Now when $|f(y)-f(x)|\leq M\>|y-x|$ for all $x$, $y\in[a,b]$ then $$\left|{f(y)-f(x)\over y-x}\right|\leq M\qquad(y\ne x)\ .$$ It follows that for each fixed $x\in[a,b]$ we have $$\left|\lim_{y\to x}{f(y)-f(x)\over y-x}\right|\leq M$$ and therefore $|f'(x)|\leq M$.

Christian Blatter
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Fix a point $x_0 \in (a,b)$ and pick an arbitrary point $x \in (a,b)$

Then

$$|f'(x) - f'(x_0)| = \lim_{w \rightarrow x,z \rightarrow x_0} |\frac{f(x)-f(w)}{w-x} - \frac{f(x_0)-f(z)}{x_0-z}| \leq 2K$$

where $K$ is the Lipschitz constant.

But then $\forall x \in (a,b)$

$$|f'(x)| \leq |f'(x) - f'(x_0)| + |f'(x_0)| \leq 2K + f'(x_0)$$

For the endpoints a similar argument will work, using monotonic limits in the above calculations.

Frank
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