Prove $f(x) = x^2sinleft(frac{1}{x}right)$ is Lipschitz (no use of derivative)
$begingroup$
Prove that $f:mathbb{R}to mathbb{R}$ such that
$$ f(x) = left{
begin{array}{c l}
x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
0 & ,quad x=0
end{array} right.$$
is Lipschitz (without use of derivatives).
Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
but I am looking for a proof, without use of derivatives.
I tried: for $x,yneq 0$:
begin{eqnarray}
x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
&=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
end{eqnarray}
so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
Since:
$$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
we get:
$$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq left(x+y+frac{y}{x}right)|x-y|.$$
Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
$+infty$, either for large $x$, or for $xapprox 0.$
Thanks in advance for the help.
real-analysis analysis lipschitz-functions
$endgroup$
add a comment |
$begingroup$
Prove that $f:mathbb{R}to mathbb{R}$ such that
$$ f(x) = left{
begin{array}{c l}
x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
0 & ,quad x=0
end{array} right.$$
is Lipschitz (without use of derivatives).
Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
but I am looking for a proof, without use of derivatives.
I tried: for $x,yneq 0$:
begin{eqnarray}
x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
&=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
end{eqnarray}
so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
Since:
$$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
we get:
$$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq left(x+y+frac{y}{x}right)|x-y|.$$
Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
$+infty$, either for large $x$, or for $xapprox 0.$
Thanks in advance for the help.
real-analysis analysis lipschitz-functions
$endgroup$
add a comment |
$begingroup$
Prove that $f:mathbb{R}to mathbb{R}$ such that
$$ f(x) = left{
begin{array}{c l}
x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
0 & ,quad x=0
end{array} right.$$
is Lipschitz (without use of derivatives).
Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
but I am looking for a proof, without use of derivatives.
I tried: for $x,yneq 0$:
begin{eqnarray}
x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
&=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
end{eqnarray}
so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
Since:
$$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
we get:
$$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq left(x+y+frac{y}{x}right)|x-y|.$$
Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
$+infty$, either for large $x$, or for $xapprox 0.$
Thanks in advance for the help.
real-analysis analysis lipschitz-functions
$endgroup$
Prove that $f:mathbb{R}to mathbb{R}$ such that
$$ f(x) = left{
begin{array}{c l}
x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
0 & ,quad x=0
end{array} right.$$
is Lipschitz (without use of derivatives).
Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
but I am looking for a proof, without use of derivatives.
I tried: for $x,yneq 0$:
begin{eqnarray}
x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
&=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
end{eqnarray}
so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
Since:
$$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
we get:
$$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq left(x+y+frac{y}{x}right)|x-y|.$$
Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
$+infty$, either for large $x$, or for $xapprox 0.$
Thanks in advance for the help.
real-analysis analysis lipschitz-functions
real-analysis analysis lipschitz-functions
asked Dec 16 '18 at 20:50
Nikolaos SkoutNikolaos Skout
2,341619
2,341619
add a comment |
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1 Answer
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$begingroup$
Writing
$$
f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
$$
left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
$$
and
$$
left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
$$
and therefore
$$
|f(x)-f(y)|leq 3|x-y| , .
$$
For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x ne 0 = y$
$$
|f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
$$
$endgroup$
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
add a comment |
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$begingroup$
Writing
$$
f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
$$
left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
$$
and
$$
left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
$$
and therefore
$$
|f(x)-f(y)|leq 3|x-y| , .
$$
For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x ne 0 = y$
$$
|f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
$$
$endgroup$
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
add a comment |
$begingroup$
Writing
$$
f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
$$
left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
$$
and
$$
left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
$$
and therefore
$$
|f(x)-f(y)|leq 3|x-y| , .
$$
For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x ne 0 = y$
$$
|f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
$$
$endgroup$
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
add a comment |
$begingroup$
Writing
$$
f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
$$
left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
$$
and
$$
left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
$$
and therefore
$$
|f(x)-f(y)|leq 3|x-y| , .
$$
For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x ne 0 = y$
$$
|f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
$$
$endgroup$
Writing
$$
f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
$$
left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
$$
and
$$
left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
$$
and therefore
$$
|f(x)-f(y)|leq 3|x-y| , .
$$
For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x ne 0 = y$
$$
|f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
$$
edited Dec 16 '18 at 21:26
answered Dec 16 '18 at 21:15
Martin RMartin R
29.3k33458
29.3k33458
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
add a comment |
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
$begingroup$
Great use of inequalities. Thank you!
$endgroup$
– Nikolaos Skout
Dec 16 '18 at 22:38
add a comment |
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