Continuous function on $[a, b]$ with bounded upper and lower derivatives on $(a, b)$ is Lipschitz. ...
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Continuous function on $[a, b]$ with bounded upper and lower derivatives on $(a, b)$ is Lipschitz.
The 2019 Stack Overflow Developer Survey Results Are InUpper semi-continuity and lower semi-continuity of particular functionsLipschitz continuity for an iterated function systemThe definition of lim sup at infinityA sequence of functions which are non-uniformly lipschitz with lipschitz limitGreatest Lower Bound of the Set of Upper Bounds for a FunctionRelation between local lipschitz constant and global lipschitz constantShowing that a function is uniformly continuous but not LipschitzApproximation of Lipschitz function in uniform normIf a sequence of unif. cont. Lipschitz functions ${f_n}$ converges unif. to an unif. cont. function f, then f is Lipschitz.Uniform convergence of Lipschitz functions and convergence of their derivatives
$begingroup$
This question was posted previously, but without the critical assumption that $f$ be continuous, and so this question was left unanswered.
I am using the following definitions:
Upper derivative of $f$ at $x$:
$overset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystylesup_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$
Lower derivative of $f$ at $x$:
$underset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystyleinf_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$.
My idea is that if, say, $overset{-}{D}$ is bounded then the $sup$ in the definition is finite call it $C$, and we can use this as the constant in the Lipschitz condition. Similarly for $underset{-}{D}$; call its bound $C'$. However, if $underset{-}{D} neq overset{-}{D}$ then $f$ is not differentiable, then these bounds are different. In this case, can I just take $textrm{max} (|C|, |C'|)$ as the Lipschitz constant?
Or am I way off?
PS: as a side question, why do we need $t$ in absolute value in the definitions? Is this because we want to approach from the left as well?
real-analysis
asked Mar 22 at 0:20
JunglemathJunglemath
1814
$endgroup$
add a comment |
$begingroup$
This question was posted previously, but without the critical assumption that $f$ be continuous, and so this question was left unanswered.
I am using the following definitions:
Upper derivative of $f$ at $x$:
$overset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystylesup_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$
Lower derivative of $f$ at $x$:
$underset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystyleinf_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$.
My idea is that if, say, $overset{-}{D}$ is bounded then the $sup$ in the definition is finite call it $C$, and we can use this as the constant in the Lipschitz condition. Similarly for $underset{-}{D}$; call its bound $C'$. However, if $underset{-}{D} neq overset{-}{D}$ then $f$ is not differentiable, then these bounds are different. In this case, can I just take $textrm{max} (|C|, |C'|)$ as the Lipschitz constant?
Or am I way off?
PS: as a side question, why do we need $t$ in absolute value in the definitions? Is this because we want to approach from the left as well?
real-analysis
asked Mar 22 at 0:20
JunglemathJunglemath
1814
$endgroup$
add a comment |
$begingroup$
This question was posted previously, but without the critical assumption that $f$ be continuous, and so this question was left unanswered.
I am using the following definitions:
Upper derivative of $f$ at $x$:
$overset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystylesup_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$
Lower derivative of $f$ at $x$:
$underset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystyleinf_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$.
My idea is that if, say, $overset{-}{D}$ is bounded then the $sup$ in the definition is finite call it $C$, and we can use this as the constant in the Lipschitz condition. Similarly for $underset{-}{D}$; call its bound $C'$. However, if $underset{-}{D} neq overset{-}{D}$ then $f$ is not differentiable, then these bounds are different. In this case, can I just take $textrm{max} (|C|, |C'|)$ as the Lipschitz constant?
Or am I way off?
PS: as a side question, why do we need $t$ in absolute value in the definitions? Is this because we want to approach from the left as well?
real-analysis
asked Mar 22 at 0:20
JunglemathJunglemath
1814
$endgroup$
This question was posted previously, but without the critical assumption that $f$ be continuous, and so this question was left unanswered.
I am using the following definitions:
Upper derivative of $f$ at $x$:
$overset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystylesup_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$
Lower derivative of $f$ at $x$:
$underset{-}{D} , f(x) = displaystyle lim_{h to 0} left [displaystyleinf_{0 < |t|leq h} frac{f(x + t) - f(x)}{t} right ]$.
My idea is that if, say, $overset{-}{D}$ is bounded then the $sup$ in the definition is finite call it $C$, and we can use this as the constant in the Lipschitz condition. Similarly for $underset{-}{D}$; call its bound $C'$. However, if $underset{-}{D} neq overset{-}{D}$ then $f$ is not differentiable, then these bounds are different. In this case, can I just take $textrm{max} (|C|, |C'|)$ as the Lipschitz constant?
Or am I way off?
PS: as a side question, why do we need $t$ in absolute value in the definitions? Is this because we want to approach from the left as well?
real-analysis
real-analysis
asked Mar 22 at 0:20
JunglemathJunglemath
1814
asked Mar 22 at 0:20
JunglemathJunglemath
1814
asked Mar 22 at 0:20
JunglemathJunglemath
1814
asked Mar 22 at 0:20
JunglemathJunglemath
1814
asked Mar 22 at 0:20
JunglemathJunglemath
1814
1814
add a comment |
add a comment |
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