Proving Local Lipschitz Property for Linear Functions on Real Numbers

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SUMMARY

The discussion confirms that the linear function \( f(x) = mx + c \) possesses both local and global Lipschitz properties on the real numbers \( \mathbb{R} \). The Lipschitz constant is established as \( m \), derived from the bounded derivative \( g'(x) = m \). It is noted that if a function is Lipschitz, it is inherently locally Lipschitz, and the proof is straightforward using the definition of Lipschitz continuity.

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  • Basic calculus concepts, particularly continuity
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onie mti
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how do i prove that f= mx+c has a local lipschitz property on R
 
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Re: locally lip function

In fact, it has the global Lipschitz property with constant $m$.
 
Re: locally lip function

Evgeny.Makarov said:
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.
 
Re: locally lip function

onie mti said:
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:
\[
|f(x_1)-f(x_2)|=|mx_1+c-(mx_2+c)|=|m(x_1-x_2)|=|m||x_1-x_2|.
\]

onie mti said:
and g' is continuous on R hence g is loc lip.
Every Lipschitz function is locally Lipschitz.
 
Re: locally lip function

Evgeny.Makarov said:
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)|
 
$x=y$ trivially implies $x\le y$.
 

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