Mixing
Differentiable only on interior points
$begingroup$
I have been pondering this question for quite a while and I would really like to get some closure on it. Whenever I have read about differentiation of functions defined on an interval, they almost always require differentiability only being valid for inner points i.e. on the interval (a,b).
Why? Can somebody thoroughly and sensibly delineate why this is the case?
One explanation I have come up with is say we attempted to define the derivative at b. In that case the left-hand and right-hand limit (the difference quotient which defines the derivative) must coincide. Points to the left of b certainly belong to the domain, however points the right of b do not belong to the domain. Therefore we can not really say anything about them. Since we cannot examine any right-hand limit, the notion of a derivative is not sensible at the boundary point.
However I am not sure the explanation entirely suffices; I am curious as to what other reasons there are to impose existence of derivatives only on interior points.
calculus
asked Feb 21 '14 at 22:31
user130591user130591
111
$endgroup$
add a comment |
$begingroup$
I have been pondering this question for quite a while and I would really like to get some closure on it. Whenever I have read about differentiation of functions defined on an interval, they almost always require differentiability only being valid for inner points i.e. on the interval (a,b).
Why? Can somebody thoroughly and sensibly delineate why this is the case?
One explanation I have come up with is say we attempted to define the derivative at b. In that case the left-hand and right-hand limit (the difference quotient which defines the derivative) must coincide. Points to the left of b certainly belong to the domain, however points the right of b do not belong to the domain. Therefore we can not really say anything about them. Since we cannot examine any right-hand limit, the notion of a derivative is not sensible at the boundary point.
However I am not sure the explanation entirely suffices; I am curious as to what other reasons there are to impose existence of derivatives only on interior points.
calculus
asked Feb 21 '14 at 22:31
user130591user130591
111
$endgroup$
1
$begingroup$
You can speak of the inesided derivative in such a case. But do you have a specific example theorem in mind? Often we need only the assumption that $f'$ exixts in the interior; that makes the theorem stronger (because it applies to more cases) compared to when we require the (onesided) derivative at the boundary to exist. Think of $sqrt x$ on $[0,1]$ for example.
$endgroup$
– Hagen von Eitzen
Feb 21 '14 at 22:34
$begingroup$
Yeah I suppose that has to be an assumption. Now that I think of it, you're right that it is stronger for the derivative to exist on an inner point. But was my explanation valid? Moreover, regarding $sqrt{x}$: The (right-hand) derivative does not even exist, so only be examining the right-hand limit we can conclude that a derivative in $x = 0$ is undefined. But there are cases where such a derivative could exist!
$endgroup$
– user130591
Feb 21 '14 at 22:56
add a comment |
$begingroup$
I have been pondering this question for quite a while and I would really like to get some closure on it. Whenever I have read about differentiation of functions defined on an interval, they almost always require differentiability only being valid for inner points i.e. on the interval (a,b).
Why? Can somebody thoroughly and sensibly delineate why this is the case?
One explanation I have come up with is say we attempted to define the derivative at b. In that case the left-hand and right-hand limit (the difference quotient which defines the derivative) must coincide. Points to the left of b certainly belong to the domain, however points the right of b do not belong to the domain. Therefore we can not really say anything about them. Since we cannot examine any right-hand limit, the notion of a derivative is not sensible at the boundary point.
However I am not sure the explanation entirely suffices; I am curious as to what other reasons there are to impose existence of derivatives only on interior points.
calculus
asked Feb 21 '14 at 22:31
user130591user130591
111
$endgroup$
I have been pondering this question for quite a while and I would really like to get some closure on it. Whenever I have read about differentiation of functions defined on an interval, they almost always require differentiability only being valid for inner points i.e. on the interval (a,b).
Why? Can somebody thoroughly and sensibly delineate why this is the case?
One explanation I have come up with is say we attempted to define the derivative at b. In that case the left-hand and right-hand limit (the difference quotient which defines the derivative) must coincide. Points to the left of b certainly belong to the domain, however points the right of b do not belong to the domain. Therefore we can not really say anything about them. Since we cannot examine any right-hand limit, the notion of a derivative is not sensible at the boundary point.
However I am not sure the explanation entirely suffices; I am curious as to what other reasons there are to impose existence of derivatives only on interior points.
calculus
calculus
asked Feb 21 '14 at 22:31
user130591user130591
111
asked Feb 21 '14 at 22:31
user130591user130591
111
asked Feb 21 '14 at 22:31
user130591user130591
111
asked Feb 21 '14 at 22:31
user130591user130591
111
asked Feb 21 '14 at 22:31
user130591user130591
111
111
1
$begingroup$
You can speak of the inesided derivative in such a case. But do you have a specific example theorem in mind? Often we need only the assumption that $f'$ exixts in the interior; that makes the theorem stronger (because it applies to more cases) compared to when we require the (onesided) derivative at the boundary to exist. Think of $sqrt x$ on $[0,1]$ for example.
$endgroup$
– Hagen von Eitzen
Feb 21 '14 at 22:34
$begingroup$
Yeah I suppose that has to be an assumption. Now that I think of it, you're right that it is stronger for the derivative to exist on an inner point. But was my explanation valid? Moreover, regarding $sqrt{x}$: The (right-hand) derivative does not even exist, so only be examining the right-hand limit we can conclude that a derivative in $x = 0$ is undefined. But there are cases where such a derivative could exist!
$endgroup$
– user130591
Feb 21 '14 at 22:56
add a comment |
1
$begingroup$
You can speak of the inesided derivative in such a case. But do you have a specific example theorem in mind? Often we need only the assumption that $f'$ exixts in the interior; that makes the theorem stronger (because it applies to more cases) compared to when we require the (onesided) derivative at the boundary to exist. Think of $sqrt x$ on $[0,1]$ for example.
$endgroup$
– Hagen von Eitzen
Feb 21 '14 at 22:34
$begingroup$
Yeah I suppose that has to be an assumption. Now that I think of it, you're right that it is stronger for the derivative to exist on an inner point. But was my explanation valid? Moreover, regarding $sqrt{x}$: The (right-hand) derivative does not even exist, so only be examining the right-hand limit we can conclude that a derivative in $x = 0$ is undefined. But there are cases where such a derivative could exist!
$endgroup$
– user130591
Feb 21 '14 at 22:56
1
1
$begingroup$
You can speak of the inesided derivative in such a case. But do you have a specific example theorem in mind? Often we need only the assumption that $f'$ exixts in the interior; that makes the theorem stronger (because it applies to more cases) compared to when we require the (onesided) derivative at the boundary to exist. Think of $sqrt x$ on $[0,1]$ for example.
$endgroup$
– Hagen von Eitzen
Feb 21 '14 at 22:34
$begingroup$
You can speak of the inesided derivative in such a case. But do you have a specific example theorem in mind? Often we need only the assumption that $f'$ exixts in the interior; that makes the theorem stronger (because it applies to more cases) compared to when we require the (onesided) derivative at the boundary to exist. Think of $sqrt x$ on $[0,1]$ for example.
$endgroup$
– Hagen von Eitzen
Feb 21 '14 at 22:34
$begingroup$
Yeah I suppose that has to be an assumption. Now that I think of it, you're right that it is stronger for the derivative to exist on an inner point. But was my explanation valid? Moreover, regarding $sqrt{x}$: The (right-hand) derivative does not even exist, so only be examining the right-hand limit we can conclude that a derivative in $x = 0$ is undefined. But there are cases where such a derivative could exist!
$endgroup$
– user130591
Feb 21 '14 at 22:56
$begingroup$
Yeah I suppose that has to be an assumption. Now that I think of it, you're right that it is stronger for the derivative to exist on an inner point. But was my explanation valid? Moreover, regarding $sqrt{x}$: The (right-hand) derivative does not even exist, so only be examining the right-hand limit we can conclude that a derivative in $x = 0$ is undefined. But there are cases where such a derivative could exist!
$endgroup$
– user130591
Feb 21 '14 at 22:56
add a comment |
1 Answer
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$begingroup$
It's just so that differentiability at a point is conserved in case you decide to expand the domain of your function.
In general, for a function $f:Arightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and $B$ is a subset of some other topological space $Y$, a limit can exist for any limit point of the domain $A$. The definition being that some $Lin{}Y$ is the limit as $x$ approaches $x_0$ of $f(x)$, i.e.
$$ lim_{xrightarrow{}x_0}f(x)=L, $$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(Ucap{}A-{x_0})subseteq{}Vcap{}B$. Notice that the limit would converge everywhere if $x_0$ was not a limit point of the domain $A$ (and also, irrelevant to this answer but also interesting, $L$ has to be in the closure of $B$).
So again, limits for a function only make sense for limit points of the function's domain. Then a function
$$F(x)=frac{f(x)-f(x_0)}{x-x_0},$$
defined in terms of the function $f$ above, would have a limit as $x$ approaches $x_0$ if $x_0$ was a limit point of $A-{x_0}$, the domain of $F$. This means that, as far as limits are concerned, the derivative could exist for any limit point of $A-{x_0}$.
But what changes when we restrict the limit points of $A-{x_0}$ to the points in the interior of $A$? Formally, the definition of
$$ lim_{xrightarrow{}x_0}F(x)=M $$
is that for all neighborhoods $V$ of $M$ there exists a neighborhood $U$ of $x_0$ such that $F(Ucap{}(A-{x_0})-{x_0})=F(Ucap{}A-{x_0})subseteq{}Vcap{}Im(F)$.
But by making $x_0$ a point in the interior of $A$ we are requiring that there exists a neighborhood $W$ of $x_0$ such that $Wsubseteq{}A$. And since $Wcap{}U$ is also a neighborhood of $x_0$ and $(Wcap{}U)cap{}A=Wcap{}U$, expanding $A$, the domain of $f$, does not change whether the limit exists or not, which makes differentiability at a point immune to expansions of the domain of the original function.
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
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add a comment |
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$begingroup$
It's just so that differentiability at a point is conserved in case you decide to expand the domain of your function.
In general, for a function $f:Arightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and $B$ is a subset of some other topological space $Y$, a limit can exist for any limit point of the domain $A$. The definition being that some $Lin{}Y$ is the limit as $x$ approaches $x_0$ of $f(x)$, i.e.
$$ lim_{xrightarrow{}x_0}f(x)=L, $$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(Ucap{}A-{x_0})subseteq{}Vcap{}B$. Notice that the limit would converge everywhere if $x_0$ was not a limit point of the domain $A$ (and also, irrelevant to this answer but also interesting, $L$ has to be in the closure of $B$).
So again, limits for a function only make sense for limit points of the function's domain. Then a function
$$F(x)=frac{f(x)-f(x_0)}{x-x_0},$$
defined in terms of the function $f$ above, would have a limit as $x$ approaches $x_0$ if $x_0$ was a limit point of $A-{x_0}$, the domain of $F$. This means that, as far as limits are concerned, the derivative could exist for any limit point of $A-{x_0}$.
But what changes when we restrict the limit points of $A-{x_0}$ to the points in the interior of $A$? Formally, the definition of
$$ lim_{xrightarrow{}x_0}F(x)=M $$
is that for all neighborhoods $V$ of $M$ there exists a neighborhood $U$ of $x_0$ such that $F(Ucap{}(A-{x_0})-{x_0})=F(Ucap{}A-{x_0})subseteq{}Vcap{}Im(F)$.
But by making $x_0$ a point in the interior of $A$ we are requiring that there exists a neighborhood $W$ of $x_0$ such that $Wsubseteq{}A$. And since $Wcap{}U$ is also a neighborhood of $x_0$ and $(Wcap{}U)cap{}A=Wcap{}U$, expanding $A$, the domain of $f$, does not change whether the limit exists or not, which makes differentiability at a point immune to expansions of the domain of the original function.
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
$endgroup$
add a comment |
$begingroup$
It's just so that differentiability at a point is conserved in case you decide to expand the domain of your function.
In general, for a function $f:Arightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and $B$ is a subset of some other topological space $Y$, a limit can exist for any limit point of the domain $A$. The definition being that some $Lin{}Y$ is the limit as $x$ approaches $x_0$ of $f(x)$, i.e.
$$ lim_{xrightarrow{}x_0}f(x)=L, $$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(Ucap{}A-{x_0})subseteq{}Vcap{}B$. Notice that the limit would converge everywhere if $x_0$ was not a limit point of the domain $A$ (and also, irrelevant to this answer but also interesting, $L$ has to be in the closure of $B$).
So again, limits for a function only make sense for limit points of the function's domain. Then a function
$$F(x)=frac{f(x)-f(x_0)}{x-x_0},$$
defined in terms of the function $f$ above, would have a limit as $x$ approaches $x_0$ if $x_0$ was a limit point of $A-{x_0}$, the domain of $F$. This means that, as far as limits are concerned, the derivative could exist for any limit point of $A-{x_0}$.
But what changes when we restrict the limit points of $A-{x_0}$ to the points in the interior of $A$? Formally, the definition of
$$ lim_{xrightarrow{}x_0}F(x)=M $$
is that for all neighborhoods $V$ of $M$ there exists a neighborhood $U$ of $x_0$ such that $F(Ucap{}(A-{x_0})-{x_0})=F(Ucap{}A-{x_0})subseteq{}Vcap{}Im(F)$.
But by making $x_0$ a point in the interior of $A$ we are requiring that there exists a neighborhood $W$ of $x_0$ such that $Wsubseteq{}A$. And since $Wcap{}U$ is also a neighborhood of $x_0$ and $(Wcap{}U)cap{}A=Wcap{}U$, expanding $A$, the domain of $f$, does not change whether the limit exists or not, which makes differentiability at a point immune to expansions of the domain of the original function.
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
$endgroup$
add a comment |
$begingroup$
It's just so that differentiability at a point is conserved in case you decide to expand the domain of your function.
In general, for a function $f:Arightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and $B$ is a subset of some other topological space $Y$, a limit can exist for any limit point of the domain $A$. The definition being that some $Lin{}Y$ is the limit as $x$ approaches $x_0$ of $f(x)$, i.e.
$$ lim_{xrightarrow{}x_0}f(x)=L, $$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(Ucap{}A-{x_0})subseteq{}Vcap{}B$. Notice that the limit would converge everywhere if $x_0$ was not a limit point of the domain $A$ (and also, irrelevant to this answer but also interesting, $L$ has to be in the closure of $B$).
So again, limits for a function only make sense for limit points of the function's domain. Then a function
$$F(x)=frac{f(x)-f(x_0)}{x-x_0},$$
defined in terms of the function $f$ above, would have a limit as $x$ approaches $x_0$ if $x_0$ was a limit point of $A-{x_0}$, the domain of $F$. This means that, as far as limits are concerned, the derivative could exist for any limit point of $A-{x_0}$.
But what changes when we restrict the limit points of $A-{x_0}$ to the points in the interior of $A$? Formally, the definition of
$$ lim_{xrightarrow{}x_0}F(x)=M $$
is that for all neighborhoods $V$ of $M$ there exists a neighborhood $U$ of $x_0$ such that $F(Ucap{}(A-{x_0})-{x_0})=F(Ucap{}A-{x_0})subseteq{}Vcap{}Im(F)$.
But by making $x_0$ a point in the interior of $A$ we are requiring that there exists a neighborhood $W$ of $x_0$ such that $Wsubseteq{}A$. And since $Wcap{}U$ is also a neighborhood of $x_0$ and $(Wcap{}U)cap{}A=Wcap{}U$, expanding $A$, the domain of $f$, does not change whether the limit exists or not, which makes differentiability at a point immune to expansions of the domain of the original function.
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
$endgroup$
It's just so that differentiability at a point is conserved in case you decide to expand the domain of your function.
In general, for a function $f:Arightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and $B$ is a subset of some other topological space $Y$, a limit can exist for any limit point of the domain $A$. The definition being that some $Lin{}Y$ is the limit as $x$ approaches $x_0$ of $f(x)$, i.e.
$$ lim_{xrightarrow{}x_0}f(x)=L, $$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(Ucap{}A-{x_0})subseteq{}Vcap{}B$. Notice that the limit would converge everywhere if $x_0$ was not a limit point of the domain $A$ (and also, irrelevant to this answer but also interesting, $L$ has to be in the closure of $B$).
So again, limits for a function only make sense for limit points of the function's domain. Then a function
$$F(x)=frac{f(x)-f(x_0)}{x-x_0},$$
defined in terms of the function $f$ above, would have a limit as $x$ approaches $x_0$ if $x_0$ was a limit point of $A-{x_0}$, the domain of $F$. This means that, as far as limits are concerned, the derivative could exist for any limit point of $A-{x_0}$.
But what changes when we restrict the limit points of $A-{x_0}$ to the points in the interior of $A$? Formally, the definition of
$$ lim_{xrightarrow{}x_0}F(x)=M $$
is that for all neighborhoods $V$ of $M$ there exists a neighborhood $U$ of $x_0$ such that $F(Ucap{}(A-{x_0})-{x_0})=F(Ucap{}A-{x_0})subseteq{}Vcap{}Im(F)$.
But by making $x_0$ a point in the interior of $A$ we are requiring that there exists a neighborhood $W$ of $x_0$ such that $Wsubseteq{}A$. And since $Wcap{}U$ is also a neighborhood of $x_0$ and $(Wcap{}U)cap{}A=Wcap{}U$, expanding $A$, the domain of $f$, does not change whether the limit exists or not, which makes differentiability at a point immune to expansions of the domain of the original function.
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
answered Nov 17 '14 at 18:55
awesomeusernameawesomeusername
657
657
add a comment |
add a comment |
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$begingroup$
You can speak of the inesided derivative in such a case. But do you have a specific example theorem in mind? Often we need only the assumption that $f'$ exixts in the interior; that makes the theorem stronger (because it applies to more cases) compared to when we require the (onesided) derivative at the boundary to exist. Think of $sqrt x$ on $[0,1]$ for example.
$endgroup$
– Hagen von Eitzen
Feb 21 '14 at 22:34
$begingroup$
Yeah I suppose that has to be an assumption. Now that I think of it, you're right that it is stronger for the derivative to exist on an inner point. But was my explanation valid? Moreover, regarding $sqrt{x}$: The (right-hand) derivative does not even exist, so only be examining the right-hand limit we can conclude that a derivative in $x = 0$ is undefined. But there are cases where such a derivative could exist!
$endgroup$
– user130591
Feb 21 '14 at 22:56