Prove or disprove sentence about $int f$Prove or disprove three sentence about $int_{a}^{b}f(t)text{d}t$...
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Prove or disprove sentence about $int f$
Prove or disprove three sentence about $int_{a}^{b}f(t)text{d}t$ (convexity)Limits of Monotone FunctionsClarification on wikipedia statement for discontinuities.Maximum and minimum values for a curveProve $2^ncdot n! ≤ (n+1)^n$ by induction.Does bounded variation imply boundednessCurve sketching for $ln(x^3 - x)$Understanding the notation $fcolon Dsubseteqmathbb{R}tomathbb{R}$ in the context of uniform continuityIntuition behind convolutionContinuous function with no local maximumProof explanation: Prove or disprove that $exists;K$ such that $|f(x)-f(y)|leq K|x-y|,;;forall; x,yin [0,1]$
$begingroup$
I post this question a few months ago. I solved the items (1) and (3), but I cannot to solve (2). Today I read this question again and I'm curious about solution of (2). Today, I had an idea, but I dont know if it works.
Idea. A monotone function has only jump discontinuities. So, $f|_{[f(x_{0}^{-}),f(x_{0}^{+})]}$ is continuous, then there is a maximum and minimum. If $w$ is a minimum on $[f(x_{0}^{-}),f(x_{0}^{+})]$, then
$$w leq f(x) Longrightarrow w(x-x_{0}) leq int_{x_{0}}^{x}f(t)dt = F(x) - F(x_{0})$$
taking, WLOG, $x geq x_{0}$. But, this works for $f$ on $[f(x_{0}^{-}),f(x_{0}^{+})]$. What about the general case?
real-analysis derivatives
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
$endgroup$
This question has an open bounty worth +50
reputation from Lucas Corrêa ending in 4 days.
This question has not received enough attention.
add a comment |
$begingroup$
I post this question a few months ago. I solved the items (1) and (3), but I cannot to solve (2). Today I read this question again and I'm curious about solution of (2). Today, I had an idea, but I dont know if it works.
Idea. A monotone function has only jump discontinuities. So, $f|_{[f(x_{0}^{-}),f(x_{0}^{+})]}$ is continuous, then there is a maximum and minimum. If $w$ is a minimum on $[f(x_{0}^{-}),f(x_{0}^{+})]$, then
$$w leq f(x) Longrightarrow w(x-x_{0}) leq int_{x_{0}}^{x}f(t)dt = F(x) - F(x_{0})$$
taking, WLOG, $x geq x_{0}$. But, this works for $f$ on $[f(x_{0}^{-}),f(x_{0}^{+})]$. What about the general case?
real-analysis derivatives
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
$endgroup$
This question has an open bounty worth +50
reputation from Lucas Corrêa ending in 4 days.
This question has not received enough attention.
1
$begingroup$
$[f(x_0^+),f(x_0^-)]$ does not necessarily belong to the domain of $f$. You cannot define the restriction of $f$ on that interval.
$endgroup$
– nicomezi
Mar 12 at 5:50
$begingroup$
@nicomezi I see. Do you have any hint? Or do you know if there is any way to adapt that idea?
$endgroup$
– Lucas Corrêa
Mar 12 at 18:22
add a comment |
$begingroup$
I post this question a few months ago. I solved the items (1) and (3), but I cannot to solve (2). Today I read this question again and I'm curious about solution of (2). Today, I had an idea, but I dont know if it works.
Idea. A monotone function has only jump discontinuities. So, $f|_{[f(x_{0}^{-}),f(x_{0}^{+})]}$ is continuous, then there is a maximum and minimum. If $w$ is a minimum on $[f(x_{0}^{-}),f(x_{0}^{+})]$, then
$$w leq f(x) Longrightarrow w(x-x_{0}) leq int_{x_{0}}^{x}f(t)dt = F(x) - F(x_{0})$$
taking, WLOG, $x geq x_{0}$. But, this works for $f$ on $[f(x_{0}^{-}),f(x_{0}^{+})]$. What about the general case?
real-analysis derivatives
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
$endgroup$
I post this question a few months ago. I solved the items (1) and (3), but I cannot to solve (2). Today I read this question again and I'm curious about solution of (2). Today, I had an idea, but I dont know if it works.
Idea. A monotone function has only jump discontinuities. So, $f|_{[f(x_{0}^{-}),f(x_{0}^{+})]}$ is continuous, then there is a maximum and minimum. If $w$ is a minimum on $[f(x_{0}^{-}),f(x_{0}^{+})]$, then
$$w leq f(x) Longrightarrow w(x-x_{0}) leq int_{x_{0}}^{x}f(t)dt = F(x) - F(x_{0})$$
taking, WLOG, $x geq x_{0}$. But, this works for $f$ on $[f(x_{0}^{-}),f(x_{0}^{+})]$. What about the general case?
real-analysis derivatives
real-analysis derivatives
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
asked Mar 12 at 5:30
Lucas CorrêaLucas Corrêa
1,4931421
1,4931421
This question has an open bounty worth +50
reputation from Lucas Corrêa ending in 4 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from Lucas Corrêa ending in 4 days.
This question has not received enough attention.
1
$begingroup$
$[f(x_0^+),f(x_0^-)]$ does not necessarily belong to the domain of $f$. You cannot define the restriction of $f$ on that interval.
$endgroup$
– nicomezi
Mar 12 at 5:50
$begingroup$
@nicomezi I see. Do you have any hint? Or do you know if there is any way to adapt that idea?
$endgroup$
– Lucas Corrêa
Mar 12 at 18:22
add a comment |
1
$begingroup$
$[f(x_0^+),f(x_0^-)]$ does not necessarily belong to the domain of $f$. You cannot define the restriction of $f$ on that interval.
$endgroup$
– nicomezi
Mar 12 at 5:50
$begingroup$
@nicomezi I see. Do you have any hint? Or do you know if there is any way to adapt that idea?
$endgroup$
– Lucas Corrêa
Mar 12 at 18:22
1
1
$begingroup$
$[f(x_0^+),f(x_0^-)]$ does not necessarily belong to the domain of $f$. You cannot define the restriction of $f$ on that interval.
$endgroup$
– nicomezi
Mar 12 at 5:50
$begingroup$
$[f(x_0^+),f(x_0^-)]$ does not necessarily belong to the domain of $f$. You cannot define the restriction of $f$ on that interval.
$endgroup$
– nicomezi
Mar 12 at 5:50
$begingroup$
@nicomezi I see. Do you have any hint? Or do you know if there is any way to adapt that idea?
$endgroup$
– Lucas Corrêa
Mar 12 at 18:22
$begingroup$
@nicomezi I see. Do you have any hint? Or do you know if there is any way to adapt that idea?
$endgroup$
– Lucas Corrêa
Mar 12 at 18:22
add a comment |
1 Answer
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$begingroup$
First suppose $x>x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt geq inf_{t in (x_0,x]} f(t) cdot (x-x_0) = f(x_0^+) cdot (x-x_0) geq w cdot (x-x_0).$$
Now let $x < x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt = - int_{x}^{x_0}f(t)dt = int_{x}^{x_0}(-f(t))dt geq inf_{t in [x,x_0)} (-f(t)) cdot (x_0-x)\
= (- sup_{t in [x,x_0)} f(t)) cdot (x_0-x) = sup_{t in [x,x_0)} f(t) cdot (x-x_0) = f(x_0^-) cdot (x-x_0) geq w cdot (x-x_0).$$
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
$endgroup$
add a comment |
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$begingroup$
First suppose $x>x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt geq inf_{t in (x_0,x]} f(t) cdot (x-x_0) = f(x_0^+) cdot (x-x_0) geq w cdot (x-x_0).$$
Now let $x < x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt = - int_{x}^{x_0}f(t)dt = int_{x}^{x_0}(-f(t))dt geq inf_{t in [x,x_0)} (-f(t)) cdot (x_0-x)\
= (- sup_{t in [x,x_0)} f(t)) cdot (x_0-x) = sup_{t in [x,x_0)} f(t) cdot (x-x_0) = f(x_0^-) cdot (x-x_0) geq w cdot (x-x_0).$$
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
$endgroup$
add a comment |
$begingroup$
First suppose $x>x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt geq inf_{t in (x_0,x]} f(t) cdot (x-x_0) = f(x_0^+) cdot (x-x_0) geq w cdot (x-x_0).$$
Now let $x < x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt = - int_{x}^{x_0}f(t)dt = int_{x}^{x_0}(-f(t))dt geq inf_{t in [x,x_0)} (-f(t)) cdot (x_0-x)\
= (- sup_{t in [x,x_0)} f(t)) cdot (x_0-x) = sup_{t in [x,x_0)} f(t) cdot (x-x_0) = f(x_0^-) cdot (x-x_0) geq w cdot (x-x_0).$$
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
$endgroup$
add a comment |
$begingroup$
First suppose $x>x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt geq inf_{t in (x_0,x]} f(t) cdot (x-x_0) = f(x_0^+) cdot (x-x_0) geq w cdot (x-x_0).$$
Now let $x < x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt = - int_{x}^{x_0}f(t)dt = int_{x}^{x_0}(-f(t))dt geq inf_{t in [x,x_0)} (-f(t)) cdot (x_0-x)\
= (- sup_{t in [x,x_0)} f(t)) cdot (x_0-x) = sup_{t in [x,x_0)} f(t) cdot (x-x_0) = f(x_0^-) cdot (x-x_0) geq w cdot (x-x_0).$$
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
$endgroup$
First suppose $x>x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt geq inf_{t in (x_0,x]} f(t) cdot (x-x_0) = f(x_0^+) cdot (x-x_0) geq w cdot (x-x_0).$$
Now let $x < x_0$. Then
$$F(x) - F(x_0) = int_{x_0}^{x}f(t)dt = - int_{x}^{x_0}f(t)dt = int_{x}^{x_0}(-f(t))dt geq inf_{t in [x,x_0)} (-f(t)) cdot (x_0-x)\
= (- sup_{t in [x,x_0)} f(t)) cdot (x_0-x) = sup_{t in [x,x_0)} f(t) cdot (x-x_0) = f(x_0^-) cdot (x-x_0) geq w cdot (x-x_0).$$
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
answered Mar 15 at 20:01
Roman HricRoman Hric
1715
1715
add a comment |
add a comment |
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1
$begingroup$
$[f(x_0^+),f(x_0^-)]$ does not necessarily belong to the domain of $f$. You cannot define the restriction of $f$ on that interval.
$endgroup$
– nicomezi
Mar 12 at 5:50
$begingroup$
@nicomezi I see. Do you have any hint? Or do you know if there is any way to adapt that idea?
$endgroup$
– Lucas Corrêa
Mar 12 at 18:22