Proof that $F$ is continuous in fundamental theorem of calculus
$begingroup$
I am studying calculus and have arrived at the first FTC. In our book it states that:
let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$
Then these properties hold:
$F_f$ is continuous on $[a,b]$
$F_f$ is differentiable on $]a,b[$
$F_f$ is a primitive for $f$
The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:
The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:
$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$
Now here is what I am not sure of:
Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?
This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?
calculus
asked Jan 4 at 22:03
ChaiChai
1276
$endgroup$
add a comment |
$begingroup$
I am studying calculus and have arrived at the first FTC. In our book it states that:
let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$
Then these properties hold:
$F_f$ is continuous on $[a,b]$
$F_f$ is differentiable on $]a,b[$
$F_f$ is a primitive for $f$
The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:
The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:
$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$
Now here is what I am not sure of:
Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?
This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?
calculus
asked Jan 4 at 22:03
ChaiChai
1276
$endgroup$
2
$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16
1
$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17
$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40
add a comment |
$begingroup$
I am studying calculus and have arrived at the first FTC. In our book it states that:
let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$
Then these properties hold:
$F_f$ is continuous on $[a,b]$
$F_f$ is differentiable on $]a,b[$
$F_f$ is a primitive for $f$
The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:
The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:
$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$
Now here is what I am not sure of:
Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?
This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?
calculus
asked Jan 4 at 22:03
ChaiChai
1276
$endgroup$
I am studying calculus and have arrived at the first FTC. In our book it states that:
let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$
Then these properties hold:
$F_f$ is continuous on $[a,b]$
$F_f$ is differentiable on $]a,b[$
$F_f$ is a primitive for $f$
The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:
The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:
$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$
Now here is what I am not sure of:
Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?
This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?
calculus
calculus
asked Jan 4 at 22:03
ChaiChai
1276
asked Jan 4 at 22:03
ChaiChai
1276
edited Jan 5 at 0:26
Chai
asked Jan 4 at 22:03
ChaiChai
1276
asked Jan 4 at 22:03
ChaiChai
1276
asked Jan 4 at 22:03
ChaiChai
1276
1276
2
$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16
1
$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17
$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40
add a comment |
2
$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16
1
$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17
$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40
2
2
$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16
$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16
1
1
$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17
$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17
$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40
$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- Right.
- Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
2
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
add a comment |
$begingroup$
$F$ is continuous if
$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$
$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$
$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$
$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$
$M = sup {|f(x)|: xin[x,y]}$
$|int_y^x f(x) dx| le |x-y|M$
$delta = frac {epsilon}{M}$
$F(x)$ is continuous.
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
- Right.
- Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
2
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
add a comment |
$begingroup$
- Right.
- Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
2
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
add a comment |
$begingroup$
- Right.
- Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
- Right.
- Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
edited Jan 5 at 9:13
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 4 at 22:18
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
2
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
add a comment |
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
2
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
$begingroup$
Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
$endgroup$
– Chai
Jan 5 at 0:56
2
2
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
$begingroup$
Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
$endgroup$
– José Carlos Santos
Jan 5 at 9:17
add a comment |
$begingroup$
$F$ is continuous if
$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$
$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$
$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$
$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$
$M = sup {|f(x)|: xin[x,y]}$
$|int_y^x f(x) dx| le |x-y|M$
$delta = frac {epsilon}{M}$
$F(x)$ is continuous.
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
$begingroup$
$F$ is continuous if
$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$
$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$
$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$
$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$
$M = sup {|f(x)|: xin[x,y]}$
$|int_y^x f(x) dx| le |x-y|M$
$delta = frac {epsilon}{M}$
$F(x)$ is continuous.
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
$begingroup$
$F$ is continuous if
$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$
$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$
$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$
$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$
$M = sup {|f(x)|: xin[x,y]}$
$|int_y^x f(x) dx| le |x-y|M$
$delta = frac {epsilon}{M}$
$F(x)$ is continuous.
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
$endgroup$
$F$ is continuous if
$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$
$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$
$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$
$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$
$M = sup {|f(x)|: xin[x,y]}$
$|int_y^x f(x) dx| le |x-y|M$
$delta = frac {epsilon}{M}$
$F(x)$ is continuous.
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
answered Jan 4 at 22:37
Doug MDoug M
44.7k31854
44.7k31854
add a comment |
add a comment |
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The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
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– Math1000
Jan 4 at 22:16
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The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
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– Math1000
Jan 4 at 22:17
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Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
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– Chai
Jan 5 at 0:40