How do I prove that if $f$ is continuous, then $F_{0}'(x) = f(x)$?
$begingroup$
If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.
Define a function $F_0(x) = int_a^x f$ for $a le x le b$
If $x$ and $x+h$ both lie in $I$, then
$F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$
= $int_x^{x+h} f = f(bar{x})h$
where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)
Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?
calculus real-analysis
asked Nov 28 '18 at 20:43
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.
Define a function $F_0(x) = int_a^x f$ for $a le x le b$
If $x$ and $x+h$ both lie in $I$, then
$F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$
= $int_x^{x+h} f = f(bar{x})h$
where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)
Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?
calculus real-analysis
asked Nov 28 '18 at 20:43
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.
Define a function $F_0(x) = int_a^x f$ for $a le x le b$
If $x$ and $x+h$ both lie in $I$, then
$F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$
= $int_x^{x+h} f = f(bar{x})h$
where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)
Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?
calculus real-analysis
asked Nov 28 '18 at 20:43
K.MK.M
693412
$endgroup$
If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.
Define a function $F_0(x) = int_a^x f$ for $a le x le b$
If $x$ and $x+h$ both lie in $I$, then
$F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$
= $int_x^{x+h} f = f(bar{x})h$
where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)
Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?
calculus real-analysis
calculus real-analysis
asked Nov 28 '18 at 20:43
K.MK.M
693412
asked Nov 28 '18 at 20:43
K.MK.M
693412
asked Nov 28 '18 at 20:43
K.MK.M
693412
asked Nov 28 '18 at 20:43
K.MK.M
693412
asked Nov 28 '18 at 20:43
K.MK.M
693412
693412
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.
So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
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$begingroup$
If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
$endgroup$
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.
So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
$endgroup$
add a comment |
$begingroup$
Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.
So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
$endgroup$
add a comment |
$begingroup$
Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.
So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
$endgroup$
Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.
So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
answered Nov 28 '18 at 20:53
Tito EliatronTito Eliatron
1,513622
1,513622
add a comment |
add a comment |
$begingroup$
If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
$endgroup$
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
add a comment |
$begingroup$
If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
$endgroup$
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
add a comment |
$begingroup$
If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
$endgroup$
If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
edited Nov 28 '18 at 21:09
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
answered Nov 28 '18 at 20:58
Will M.Will M.
2,630315
2,630315
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
add a comment |
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
where does the $(f -f(x)) + f(x)h$ come from?
$endgroup$
– K.M
Nov 28 '18 at 21:06
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
$begingroup$
Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
$endgroup$
– Will M.
Nov 28 '18 at 21:09
add a comment |
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