Not continuous on endpoints and differentiability The 2019 Stack Overflow Developer Survey Results Are InAnalysis-Baby Rudin's differentiability and continuity: theorem 5.2 and 5.6Directly proving continuous differentiabilityWhy differentiability implies continuity, but continuity does not imply differentiability?Continuity is required for differentiability?Differentiability implies continuous derivative?Does differentiability on a set imply continuous differentiability on the set? Counterexample?Why is continuity permissible at endpoints but not differentiability?How to avoid ambiguity defining continuity / differentiability of multivariable functionUniform Lipshitz continuity implies Continuous DifferentiabilityTopologically, is there a definition of differentiability that is dependent on the underlying topology, similar to continuity?
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Not continuous on endpoints and differentiability
The 2019 Stack Overflow Developer Survey Results Are InAnalysis-Baby Rudin's differentiability and continuity: theorem 5.2 and 5.6Directly proving continuous differentiabilityWhy differentiability implies continuity, but continuity does not imply differentiability?Continuity is required for differentiability?Differentiability implies continuous derivative?Does differentiability on a set imply continuous differentiability on the set? Counterexample?Why is continuity permissible at endpoints but not differentiability?How to avoid ambiguity defining continuity / differentiability of multivariable functionUniform Lipshitz continuity implies Continuous DifferentiabilityTopologically, is there a definition of differentiability that is dependent on the underlying topology, similar to continuity?
$begingroup$
I have a general question.
Say $f$ is a real function from $(a,b)$ to $mathbb R$.
We usually prove continuity on $[a,b]$, but if f were continuous on the open interval $(a,b)$, would there be any issues with differentiability? Do we need continuity on $[a,b]$ in order to have differentiability on (a,b)?
real-analysis derivatives continuity
edited Apr 7 at 18:06
Haris Gusic
3,531627
asked Apr 7 at 17:55
JessJess
257
$endgroup$
add a comment |
$begingroup$
I have a general question.
Say $f$ is a real function from $(a,b)$ to $mathbb R$.
We usually prove continuity on $[a,b]$, but if f were continuous on the open interval $(a,b)$, would there be any issues with differentiability? Do we need continuity on $[a,b]$ in order to have differentiability on (a,b)?
real-analysis derivatives continuity
edited Apr 7 at 18:06
Haris Gusic
3,531627
asked Apr 7 at 17:55
JessJess
257
$endgroup$
1
$begingroup$
We don’t even need $f$ to be defined at $a$ or at $b$ for $f$ to be differentiable on $(a,b)$. The reason you will see a lot of theorems that ask for functions that are “continuous on $[a,b]$ and differentiable on $(a,b)$” is that they need continuity at the endpoints, but don’t need differentiability there; that is, because they are asking for less than they would be asking if they just said “differentiable on $[a,b]$”.
$endgroup$
– Arturo Magidin
Apr 7 at 18:00
add a comment |
$begingroup$
I have a general question.
Say $f$ is a real function from $(a,b)$ to $mathbb R$.
We usually prove continuity on $[a,b]$, but if f were continuous on the open interval $(a,b)$, would there be any issues with differentiability? Do we need continuity on $[a,b]$ in order to have differentiability on (a,b)?
real-analysis derivatives continuity
edited Apr 7 at 18:06
Haris Gusic
3,531627
asked Apr 7 at 17:55
JessJess
257
$endgroup$
I have a general question.
Say $f$ is a real function from $(a,b)$ to $mathbb R$.
We usually prove continuity on $[a,b]$, but if f were continuous on the open interval $(a,b)$, would there be any issues with differentiability? Do we need continuity on $[a,b]$ in order to have differentiability on (a,b)?
real-analysis derivatives continuity
real-analysis derivatives continuity
edited Apr 7 at 18:06
Haris Gusic
3,531627
asked Apr 7 at 17:55
JessJess
257
edited Apr 7 at 18:06
Haris Gusic
3,531627
asked Apr 7 at 17:55
JessJess
257
edited Apr 7 at 18:06
Haris Gusic
3,531627
edited Apr 7 at 18:06
Haris Gusic
3,531627
edited Apr 7 at 18:06
Haris Gusic
3,531627
3,531627
asked Apr 7 at 17:55
JessJess
257
asked Apr 7 at 17:55
JessJess
257
asked Apr 7 at 17:55
JessJess
257
257
1
$begingroup$
We don’t even need $f$ to be defined at $a$ or at $b$ for $f$ to be differentiable on $(a,b)$. The reason you will see a lot of theorems that ask for functions that are “continuous on $[a,b]$ and differentiable on $(a,b)$” is that they need continuity at the endpoints, but don’t need differentiability there; that is, because they are asking for less than they would be asking if they just said “differentiable on $[a,b]$”.
$endgroup$
– Arturo Magidin
Apr 7 at 18:00
add a comment |
1
$begingroup$
We don’t even need $f$ to be defined at $a$ or at $b$ for $f$ to be differentiable on $(a,b)$. The reason you will see a lot of theorems that ask for functions that are “continuous on $[a,b]$ and differentiable on $(a,b)$” is that they need continuity at the endpoints, but don’t need differentiability there; that is, because they are asking for less than they would be asking if they just said “differentiable on $[a,b]$”.
$endgroup$
– Arturo Magidin
Apr 7 at 18:00
1
1
$begingroup$
We don’t even need $f$ to be defined at $a$ or at $b$ for $f$ to be differentiable on $(a,b)$. The reason you will see a lot of theorems that ask for functions that are “continuous on $[a,b]$ and differentiable on $(a,b)$” is that they need continuity at the endpoints, but don’t need differentiability there; that is, because they are asking for less than they would be asking if they just said “differentiable on $[a,b]$”.
$endgroup$
– Arturo Magidin
Apr 7 at 18:00
$begingroup$
We don’t even need $f$ to be defined at $a$ or at $b$ for $f$ to be differentiable on $(a,b)$. The reason you will see a lot of theorems that ask for functions that are “continuous on $[a,b]$ and differentiable on $(a,b)$” is that they need continuity at the endpoints, but don’t need differentiability there; that is, because they are asking for less than they would be asking if they just said “differentiable on $[a,b]$”.
$endgroup$
– Arturo Magidin
Apr 7 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. Let's start from the claim that says that differentiability at a point implies continuity at that point. Differentiability on an interval $(a,b)$ is equivalent with differentiability at each point of that interval. This implies that the function is continuous at each point of that interval, i.e. continuous on the interval $(a,b)$. This says nothing about the endpoints.
As an example, consider the function $tan(x)$. It is continuous on $(pi/2,pi/2)$, but not on $[-pi/2,pi/2]$ (it isn't even defined at the endpoints). Still, it is differentiable on $(-pi/2, pi/2)$.
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
$endgroup$
add a comment |
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$begingroup$
No. Let's start from the claim that says that differentiability at a point implies continuity at that point. Differentiability on an interval $(a,b)$ is equivalent with differentiability at each point of that interval. This implies that the function is continuous at each point of that interval, i.e. continuous on the interval $(a,b)$. This says nothing about the endpoints.
As an example, consider the function $tan(x)$. It is continuous on $(pi/2,pi/2)$, but not on $[-pi/2,pi/2]$ (it isn't even defined at the endpoints). Still, it is differentiable on $(-pi/2, pi/2)$.
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
$endgroup$
add a comment |
$begingroup$
No. Let's start from the claim that says that differentiability at a point implies continuity at that point. Differentiability on an interval $(a,b)$ is equivalent with differentiability at each point of that interval. This implies that the function is continuous at each point of that interval, i.e. continuous on the interval $(a,b)$. This says nothing about the endpoints.
As an example, consider the function $tan(x)$. It is continuous on $(pi/2,pi/2)$, but not on $[-pi/2,pi/2]$ (it isn't even defined at the endpoints). Still, it is differentiable on $(-pi/2, pi/2)$.
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
$endgroup$
add a comment |
$begingroup$
No. Let's start from the claim that says that differentiability at a point implies continuity at that point. Differentiability on an interval $(a,b)$ is equivalent with differentiability at each point of that interval. This implies that the function is continuous at each point of that interval, i.e. continuous on the interval $(a,b)$. This says nothing about the endpoints.
As an example, consider the function $tan(x)$. It is continuous on $(pi/2,pi/2)$, but not on $[-pi/2,pi/2]$ (it isn't even defined at the endpoints). Still, it is differentiable on $(-pi/2, pi/2)$.
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
$endgroup$
No. Let's start from the claim that says that differentiability at a point implies continuity at that point. Differentiability on an interval $(a,b)$ is equivalent with differentiability at each point of that interval. This implies that the function is continuous at each point of that interval, i.e. continuous on the interval $(a,b)$. This says nothing about the endpoints.
As an example, consider the function $tan(x)$. It is continuous on $(pi/2,pi/2)$, but not on $[-pi/2,pi/2]$ (it isn't even defined at the endpoints). Still, it is differentiable on $(-pi/2, pi/2)$.
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
edited Apr 7 at 18:10
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
answered Apr 7 at 17:57
Haris GusicHaris Gusic
3,531627
3,531627
add a comment |
add a comment |
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$begingroup$
We don’t even need $f$ to be defined at $a$ or at $b$ for $f$ to be differentiable on $(a,b)$. The reason you will see a lot of theorems that ask for functions that are “continuous on $[a,b]$ and differentiable on $(a,b)$” is that they need continuity at the endpoints, but don’t need differentiability there; that is, because they are asking for less than they would be asking if they just said “differentiable on $[a,b]$”.
$endgroup$
– Arturo Magidin
Apr 7 at 18:00