Prove that if $g(x)leq g(y)$ then $F(x)leq F(y),$ where $F(x)=int^{g(x)}_{a}f(t)dt$












1












$begingroup$


Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24


















1












$begingroup$


Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24
















1












1








1


1



$begingroup$


Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification







share|cite|improve this question











$endgroup$




Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification




real-analysis calculus analysis proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 3:27







Omojola Micheal

















asked Jan 14 at 21:04









Omojola MichealOmojola Micheal

1,851324




1,851324








  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24
















  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24










2




2




$begingroup$
Looks OK, but the best you can conclude is that $F$ is nondecreasing.
$endgroup$
– Umberto P.
Jan 14 at 21:09




$begingroup$
Looks OK, but the best you can conclude is that $F$ is nondecreasing.
$endgroup$
– Umberto P.
Jan 14 at 21:09












$begingroup$
Thanks for your feedback! I'll edit!
$endgroup$
– Omojola Micheal
Jan 14 at 21:10






$begingroup$
Thanks for your feedback! I'll edit!
$endgroup$
– Omojola Micheal
Jan 14 at 21:10






1




1




$begingroup$
No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
$endgroup$
– Arthur
Jan 14 at 21:14






$begingroup$
No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
$endgroup$
– Arthur
Jan 14 at 21:14














$begingroup$
@Arthur: Hmm... I see you what you've stated!
$endgroup$
– Omojola Micheal
Jan 14 at 21:18




$begingroup$
@Arthur: Hmm... I see you what you've stated!
$endgroup$
– Omojola Micheal
Jan 14 at 21:18




1




1




$begingroup$
@UmbertoP. I think the constant function is monotonically increasing as well.
$endgroup$
– Botond
Jan 14 at 21:24






$begingroup$
@UmbertoP. I think the constant function is monotonically increasing as well.
$endgroup$
– Botond
Jan 14 at 21:24












1 Answer
1






active

oldest

votes


















2












$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07











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1 Answer
1






active

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active

oldest

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active

oldest

votes









2












$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07
















2












$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07














2












2








2





$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$



This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 21:59

























answered Jan 14 at 21:47









Maximilian JanischMaximilian Janisch

44110




44110












  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07


















  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07
















$begingroup$
Maximilian Janisch: This is nice! (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 22:07




$begingroup$
Maximilian Janisch: This is nice! (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 22:07


















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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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