About positive part of difference fuctions.
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What can we Say that $(f-g)^{+}$? $(f-g)^{+}=(f^{+}+(-g)^{+})=f^{+}-g^{-}$?
real-analysis calculus functional-analysis
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
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add a comment |
$begingroup$
What can we Say that $(f-g)^{+}$? $(f-g)^{+}=(f^{+}+(-g)^{+})=f^{+}-g^{-}$?
real-analysis calculus functional-analysis
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
What can we Say that $(f-g)^{+}$? $(f-g)^{+}=(f^{+}+(-g)^{+})=f^{+}-g^{-}$?
real-analysis calculus functional-analysis
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
$endgroup$
What can we Say that $(f-g)^{+}$? $(f-g)^{+}=(f^{+}+(-g)^{+})=f^{+}-g^{-}$?
real-analysis calculus functional-analysis
real-analysis calculus functional-analysis
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
asked Dec 18 '18 at 15:50
eraldcoileraldcoil
395211
395211
add a comment |
add a comment |
1 Answer
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The positive and the negative part of a function are non negative function. In general it is not true that $ (-g)^+ = - g^-$. It is true if and only if $g=0$.
But we know (see for instance https://en.wikipedia.org/wiki/Positive_and_negative_parts) that for every function
$$ f^+ = frac{ | f | +f }{2} quad f^- = frac{ | f | -f }{2}.$$
Thus
$$(f-g)^+ = frac{|f-g| + f-g }{2} leq frac{|f| + |g| + f-g }{2} = frac{|f| + f}{2} + frac{|g| -g }{2} = f^+ + (-g)^+.$$
answered Dec 18 '18 at 17:10
HermioneHermione
19619
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1 Answer
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1 Answer
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$begingroup$
The positive and the negative part of a function are non negative function. In general it is not true that $ (-g)^+ = - g^-$. It is true if and only if $g=0$.
But we know (see for instance https://en.wikipedia.org/wiki/Positive_and_negative_parts) that for every function
$$ f^+ = frac{ | f | +f }{2} quad f^- = frac{ | f | -f }{2}.$$
Thus
$$(f-g)^+ = frac{|f-g| + f-g }{2} leq frac{|f| + |g| + f-g }{2} = frac{|f| + f}{2} + frac{|g| -g }{2} = f^+ + (-g)^+.$$
answered Dec 18 '18 at 17:10
HermioneHermione
19619
$endgroup$
add a comment |
$begingroup$
The positive and the negative part of a function are non negative function. In general it is not true that $ (-g)^+ = - g^-$. It is true if and only if $g=0$.
But we know (see for instance https://en.wikipedia.org/wiki/Positive_and_negative_parts) that for every function
$$ f^+ = frac{ | f | +f }{2} quad f^- = frac{ | f | -f }{2}.$$
Thus
$$(f-g)^+ = frac{|f-g| + f-g }{2} leq frac{|f| + |g| + f-g }{2} = frac{|f| + f}{2} + frac{|g| -g }{2} = f^+ + (-g)^+.$$
answered Dec 18 '18 at 17:10
HermioneHermione
19619
$endgroup$
add a comment |
$begingroup$
The positive and the negative part of a function are non negative function. In general it is not true that $ (-g)^+ = - g^-$. It is true if and only if $g=0$.
But we know (see for instance https://en.wikipedia.org/wiki/Positive_and_negative_parts) that for every function
$$ f^+ = frac{ | f | +f }{2} quad f^- = frac{ | f | -f }{2}.$$
Thus
$$(f-g)^+ = frac{|f-g| + f-g }{2} leq frac{|f| + |g| + f-g }{2} = frac{|f| + f}{2} + frac{|g| -g }{2} = f^+ + (-g)^+.$$
answered Dec 18 '18 at 17:10
HermioneHermione
19619
$endgroup$
The positive and the negative part of a function are non negative function. In general it is not true that $ (-g)^+ = - g^-$. It is true if and only if $g=0$.
But we know (see for instance https://en.wikipedia.org/wiki/Positive_and_negative_parts) that for every function
$$ f^+ = frac{ | f | +f }{2} quad f^- = frac{ | f | -f }{2}.$$
Thus
$$(f-g)^+ = frac{|f-g| + f-g }{2} leq frac{|f| + |g| + f-g }{2} = frac{|f| + f}{2} + frac{|g| -g }{2} = f^+ + (-g)^+.$$
answered Dec 18 '18 at 17:10
HermioneHermione
19619
answered Dec 18 '18 at 17:10
HermioneHermione
19619
answered Dec 18 '18 at 17:10
HermioneHermione
19619
answered Dec 18 '18 at 17:10
HermioneHermione
19619
19619
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