is this function regular?
$begingroup$
I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.
A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.
The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
$$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
$$begin{align}
f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
&=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
end{align}$$
The definition of regular is not straightforward for beginners. Let's use an example to study it.
Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
Then is function $f_1$ regular at $x=pm 1$ or
not?
Here is my calculation.
$$
begin{align}
f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
&=begin{cases}0,&v>0,\2v,&v<0.end{cases}
end{align}
$$
and
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
&=begin{cases}
2v,&v>0,\
0,&v<0.
end{cases}
end{align}$$
Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).
More calculation steps:
Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
$$begin{align}
f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
&=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
&=begin{cases}
sup&begin{cases}
0,&k>0,\
begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
end{cases},v>0,\
sup&begin{cases}
2v,&k<0,\
begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
end{cases},v<0,
end{cases}\
&=begin{cases}2v,&v>0,\0,&v<0end{cases}
end{align}$$
I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
if $v>0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
&=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
&=0.
end{align}$$
And if $v<0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
&=2v
end{align}$$
The original version in the paper are
the piecewise potential function and regularity analysis.
Obviously, there is something wrong. Can anyone point it out?
Thank you.
real-analysis derivatives control-theory
$endgroup$
add a comment |
$begingroup$
I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.
A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.
The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
$$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
$$begin{align}
f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
&=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
end{align}$$
The definition of regular is not straightforward for beginners. Let's use an example to study it.
Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
Then is function $f_1$ regular at $x=pm 1$ or
not?
Here is my calculation.
$$
begin{align}
f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
&=begin{cases}0,&v>0,\2v,&v<0.end{cases}
end{align}
$$
and
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
&=begin{cases}
2v,&v>0,\
0,&v<0.
end{cases}
end{align}$$
Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).
More calculation steps:
Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
$$begin{align}
f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
&=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
&=begin{cases}
sup&begin{cases}
0,&k>0,\
begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
end{cases},v>0,\
sup&begin{cases}
2v,&k<0,\
begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
end{cases},v<0,
end{cases}\
&=begin{cases}2v,&v>0,\0,&v<0end{cases}
end{align}$$
I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
if $v>0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
&=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
&=0.
end{align}$$
And if $v<0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
&=2v
end{align}$$
The original version in the paper are
the piecewise potential function and regularity analysis.
Obviously, there is something wrong. Can anyone point it out?
Thank you.
real-analysis derivatives control-theory
$endgroup$
add a comment |
$begingroup$
I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.
A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.
The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
$$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
$$begin{align}
f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
&=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
end{align}$$
The definition of regular is not straightforward for beginners. Let's use an example to study it.
Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
Then is function $f_1$ regular at $x=pm 1$ or
not?
Here is my calculation.
$$
begin{align}
f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
&=begin{cases}0,&v>0,\2v,&v<0.end{cases}
end{align}
$$
and
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
&=begin{cases}
2v,&v>0,\
0,&v<0.
end{cases}
end{align}$$
Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).
More calculation steps:
Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
$$begin{align}
f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
&=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
&=begin{cases}
sup&begin{cases}
0,&k>0,\
begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
end{cases},v>0,\
sup&begin{cases}
2v,&k<0,\
begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
end{cases},v<0,
end{cases}\
&=begin{cases}2v,&v>0,\0,&v<0end{cases}
end{align}$$
I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
if $v>0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
&=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
&=0.
end{align}$$
And if $v<0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
&=2v
end{align}$$
The original version in the paper are
the piecewise potential function and regularity analysis.
Obviously, there is something wrong. Can anyone point it out?
Thank you.
real-analysis derivatives control-theory
$endgroup$
I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.
A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.
The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
$$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
$$begin{align}
f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
&=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
end{align}$$
The definition of regular is not straightforward for beginners. Let's use an example to study it.
Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
Then is function $f_1$ regular at $x=pm 1$ or
not?
Here is my calculation.
$$
begin{align}
f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
&=begin{cases}0,&v>0,\2v,&v<0.end{cases}
end{align}
$$
and
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
&=begin{cases}
2v,&v>0,\
0,&v<0.
end{cases}
end{align}$$
Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).
More calculation steps:
Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
$$begin{align}
f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
&=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
&=begin{cases}
sup&begin{cases}
0,&k>0,\
begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
end{cases},v>0,\
sup&begin{cases}
2v,&k<0,\
begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
end{cases},v<0,
end{cases}\
&=begin{cases}2v,&v>0,\0,&v<0end{cases}
end{align}$$
I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
if $v>0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
&=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
&=0.
end{align}$$
And if $v<0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
&=2v
end{align}$$
The original version in the paper are
the piecewise potential function and regularity analysis.
Obviously, there is something wrong. Can anyone point it out?
Thank you.
real-analysis derivatives control-theory
real-analysis derivatives control-theory
edited Jan 9 at 16:41
wjxjmj
asked Jan 9 at 8:13
wjxjmjwjxjmj
62
62
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