A function f:R→R is defined to be Lipschitz if there is a constant $K>0$ such that for all a,b∈R $|f(a)−f(b)|≤K|a−b$| (a) Use the mean value theorem to show that the function $f(x)=2sin(x)$ is Lipschitz.

(b) Suppose $f:R→R$ is Lipschitz. Prove that f is continuous. (Hints: Use squeeze law and/or the epsilon-delta definition of limit.

--My solution:

Part a:

$2sin(x)$ is continuous on $R$, in particular on $[b,a]$ and differentiable on $(b,a)$. By MVT, there exists $c∈(b,a)$ such that $$f′(c)= \frac{f(a)−f(b)}{a−b} $$ But $f′(x)=2cos(x)$, therefore $$f′(c)=2cos(c)$$ therefore $$2cos(c)= \frac{f(a)−f(b)}{a−b}$$ therefore

$$|2cos(c)|= \frac{|f(a)−f(b)|}{|a−b|}$$

but $|2cos(c)| ≤ 2$, therefore $$\frac{|f(a)−f(b)|}{|a−b|} ≤ 2$$

i.e. $$|f(a)-f(b)| ≤ 2|a-b|$$

Part b:

let $ϵ > 0$ and $x$, $x_0$ $∈ R$

for $f$ to be continuous at $x_0$, need to ensure that $|f(x) - f(x_0)| < ϵ$

now from part (a) we have: $$|f(x) - f(x_0)| ≤ K|x - x_0|$$

when $|x - x_0| < δ$ we have: $$|f(x) - f(x_0)| ≤ Kδ$$

let $δ = \frac{ϵ}{K}$

therefore we have: $$|f(x) - f(x_0)| < ϵ$$

therefore $f$ is continuous on $R$

Could someone please verify that this proof is rigorous enough and if not, point out where it needs to be adjusted. Thanks in advance!

  • $\begingroup$ Looks good to me, nice job. $\endgroup$ – Joseph G. May 12 '13 at 1:23

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