MHB Proving Local Lipschitz Property for Linear Functions on Real Numbers

[onie mti]

how do i prove that f= mx+c has a local lipschitz property on R

 

[Evgeny.Makarov]

Re: locally lip function

In fact, it has the global Lipschitz property with constant $m$.

 

[onie mti]

Re: locally lip function

In fact, it has the global Lipschitz property with constant $m$.

is it acceptable to say;
suppose that g is differentiable on R.
g'(x)= m
If the derivative is bounded on R, then g is Lip on R and any upper bound for |g'(x)|=m is the lip constant.

and g' is continuous on R hence g is loc lip.

 

[Evgeny.Makarov]

Re: locally lip function

is it acceptable to say;
suppose that g is differentiable on R.
g'(x)= m
If the derivative is bounded on R, then g is Lip on R and any upper bound for |g'(x)|=m is the lip constant.

Yes. Of course, proving that $f(x)=mx+c$ is Lipschitz by definition is also easy:
\[
|f(x_1)-f(x_2)|=|mx_1+c-(mx_2+c)|=|m(x_1-x_2)|=|m||x_1-x_2|.
\]

and g' is continuous on R hence g is loc lip.

Every Lipschitz function is locally Lipschitz.

 

[onie mti]

Re: locally lip function

Yes. Of course, proving that $f(x)=mx+c$ is Lipschitz by definition is also easy:
\[
|f(x_1)-f(x_2)|=|mx_1+c-(mx_2+c)|=|m(x_1-x_2)|=|m||x_1-x_2|.
\]

Every Lipschitz function is locally Lipschitz.

but doesn't the def of a Lip function say: | f(x_1) - f(x_2)| less than equal m|(x_1) -(x_2)|

 

[Evgeny.Makarov]

$x=y$ trivially implies $x\le y$.