Mixing
Can the non-differentiability of a function $f:R^n to R$ always be proved by using directional derivative?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
$F (x ,y) = |x| + |y|$ when $xy neq 0$
and $F(x,y) =0 $ elsewhere.
How can I prove or disprove this function is differentiable at $(0,0)$?
My Try : The directional derivative at $(0,0)$ in the direction $(h,k)$ is $|h| + |k|$ (where $hk neq 0$ )which is not a linear function of $(h,k)$. So it can not have differentiation at $(0,0)$.
Can anyone please tell me if I am wrong?
calculus real-analysis derivatives partial-derivative
asked Jul 24 at 5:17
cmi
6319
add a comment |Â
up vote
0
down vote
favorite
$F (x ,y) = |x| + |y|$ when $xy neq 0$
and $F(x,y) =0 $ elsewhere.
How can I prove or disprove this function is differentiable at $(0,0)$?
My Try : The directional derivative at $(0,0)$ in the direction $(h,k)$ is $|h| + |k|$ (where $hk neq 0$ )which is not a linear function of $(h,k)$. So it can not have differentiation at $(0,0)$.
Can anyone please tell me if I am wrong?
calculus real-analysis derivatives partial-derivative
asked Jul 24 at 5:17
cmi
6319
The question in the title is completely different from the question in the post
– Jack M
Jul 24 at 7:40
Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_v+wf = D_vf + D_wf$ for all $v,w$ (note that $D_lambda vf = lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature.
– Paul Frost
Jul 24 at 8:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$F (x ,y) = |x| + |y|$ when $xy neq 0$
and $F(x,y) =0 $ elsewhere.
How can I prove or disprove this function is differentiable at $(0,0)$?
My Try : The directional derivative at $(0,0)$ in the direction $(h,k)$ is $|h| + |k|$ (where $hk neq 0$ )which is not a linear function of $(h,k)$. So it can not have differentiation at $(0,0)$.
Can anyone please tell me if I am wrong?
calculus real-analysis derivatives partial-derivative
asked Jul 24 at 5:17
cmi
6319
$F (x ,y) = |x| + |y|$ when $xy neq 0$
and $F(x,y) =0 $ elsewhere.
How can I prove or disprove this function is differentiable at $(0,0)$?
My Try : The directional derivative at $(0,0)$ in the direction $(h,k)$ is $|h| + |k|$ (where $hk neq 0$ )which is not a linear function of $(h,k)$. So it can not have differentiation at $(0,0)$.
Can anyone please tell me if I am wrong?
calculus real-analysis derivatives partial-derivative
asked Jul 24 at 5:17
cmi
6319
edited Jul 24 at 6:46
asked Jul 24 at 5:17
cmi
6319
asked Jul 24 at 5:17
cmi
6319
asked Jul 24 at 5:17
cmi
6319
6319
The question in the title is completely different from the question in the post
– Jack M
Jul 24 at 7:40
Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_v+wf = D_vf + D_wf$ for all $v,w$ (note that $D_lambda vf = lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature.
– Paul Frost
Jul 24 at 8:03
add a comment |Â
The question in the title is completely different from the question in the post
– Jack M
Jul 24 at 7:40
Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_v+wf = D_vf + D_wf$ for all $v,w$ (note that $D_lambda vf = lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature.
– Paul Frost
Jul 24 at 8:03
The question in the title is completely different from the question in the post
– Jack M
Jul 24 at 7:40
The question in the title is completely different from the question in the post
– Jack M
Jul 24 at 7:40
Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_v+wf = D_vf + D_wf$ for all $v,w$ (note that $D_lambda vf = lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature.
– Paul Frost
Jul 24 at 8:03
Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_v+wf = D_vf + D_wf$ for all $v,w$ (note that $D_lambda vf = lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature.
– Paul Frost
Jul 24 at 8:03
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
That's a fine approach. In order to be differentiable, the directional derivative must be linear, which is not the case for this function.
The only nitpick I have is with the directional derivative you've computed. It's right when $h$ and $k$ is non-zero, but the function is constant along the axes, so the directional derivatives will be $0$ whenever $(h, k)$ points along the axes.
But, either way, it's definitely not linear.
answered Jul 24 at 5:33
Theo Bendit
12k1843
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
add a comment |Â
up vote
2
down vote
For the question in the title, no. If $f(x)=1$ if $0<x=y^2$ and zero otherwise, the directional derivative at 0 in any direction is 0, but the function is not even continuous at zero.
answered Jul 24 at 5:46
Calvin Khor
8,10911133
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
add a comment |Â
up vote
0
down vote
The question in the title asks, at least in my understandig, also whether there are functions for which no directional derivatives exist at all.
In dead such (continuous) functions exist. Take any continuous nowhere differentiable function $gcolonmathbbRtomathbbR$. Then $fcolonmathbbR^2tomathbbR$ with $f(x,y):=g(x)+g(y)$ has the properties to be continuous and to have no directional derivatives at all.
Edit: I'm rather sure that my claim is true, but it seems to be not so easy to proof it, as the fact that two limits do not exist does not always imply that the limit of the sum also does not exist.
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
That's a fine approach. In order to be differentiable, the directional derivative must be linear, which is not the case for this function.
The only nitpick I have is with the directional derivative you've computed. It's right when $h$ and $k$ is non-zero, but the function is constant along the axes, so the directional derivatives will be $0$ whenever $(h, k)$ points along the axes.
But, either way, it's definitely not linear.
answered Jul 24 at 5:33
Theo Bendit
12k1843
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
add a comment |Â
up vote
2
down vote
accepted
That's a fine approach. In order to be differentiable, the directional derivative must be linear, which is not the case for this function.
The only nitpick I have is with the directional derivative you've computed. It's right when $h$ and $k$ is non-zero, but the function is constant along the axes, so the directional derivatives will be $0$ whenever $(h, k)$ points along the axes.
But, either way, it's definitely not linear.
answered Jul 24 at 5:33
Theo Bendit
12k1843
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
That's a fine approach. In order to be differentiable, the directional derivative must be linear, which is not the case for this function.
The only nitpick I have is with the directional derivative you've computed. It's right when $h$ and $k$ is non-zero, but the function is constant along the axes, so the directional derivatives will be $0$ whenever $(h, k)$ points along the axes.
But, either way, it's definitely not linear.
answered Jul 24 at 5:33
Theo Bendit
12k1843
That's a fine approach. In order to be differentiable, the directional derivative must be linear, which is not the case for this function.
The only nitpick I have is with the directional derivative you've computed. It's right when $h$ and $k$ is non-zero, but the function is constant along the axes, so the directional derivatives will be $0$ whenever $(h, k)$ points along the axes.
But, either way, it's definitely not linear.
answered Jul 24 at 5:33
Theo Bendit
12k1843
answered Jul 24 at 5:33
Theo Bendit
12k1843
answered Jul 24 at 5:33
Theo Bendit
12k1843
answered Jul 24 at 5:33
Theo Bendit
12k1843
12k1843
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
add a comment |Â
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit
– cmi
Jul 24 at 6:19
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
@cmi You absolutely can. If you put in the caveat about $hk neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true.
– Theo Bendit
Jul 24 at 6:22
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
I have edited..Is it now okay?@Theo Bendit
– cmi
Jul 24 at 6:47
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
@cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-)
– Theo Bendit
Jul 24 at 6:49
add a comment |Â
up vote
2
down vote
For the question in the title, no. If $f(x)=1$ if $0<x=y^2$ and zero otherwise, the directional derivative at 0 in any direction is 0, but the function is not even continuous at zero.
answered Jul 24 at 5:46
Calvin Khor
8,10911133
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
add a comment |Â
up vote
2
down vote
For the question in the title, no. If $f(x)=1$ if $0<x=y^2$ and zero otherwise, the directional derivative at 0 in any direction is 0, but the function is not even continuous at zero.
answered Jul 24 at 5:46
Calvin Khor
8,10911133
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For the question in the title, no. If $f(x)=1$ if $0<x=y^2$ and zero otherwise, the directional derivative at 0 in any direction is 0, but the function is not even continuous at zero.
answered Jul 24 at 5:46
Calvin Khor
8,10911133
For the question in the title, no. If $f(x)=1$ if $0<x=y^2$ and zero otherwise, the directional derivative at 0 in any direction is 0, but the function is not even continuous at zero.
answered Jul 24 at 5:46
Calvin Khor
8,10911133
answered Jul 24 at 5:46
Calvin Khor
8,10911133
answered Jul 24 at 5:46
Calvin Khor
8,10911133
answered Jul 24 at 5:46
Calvin Khor
8,10911133
8,10911133
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
add a comment |Â
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor
– cmi
Jul 24 at 6:40
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
@cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making)
– Calvin Khor
Jul 24 at 6:41
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor
– cmi
Jul 24 at 6:43
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear
– Calvin Khor
Jul 24 at 6:48
add a comment |Â
up vote
0
down vote
The question in the title asks, at least in my understandig, also whether there are functions for which no directional derivatives exist at all.
In dead such (continuous) functions exist. Take any continuous nowhere differentiable function $gcolonmathbbRtomathbbR$. Then $fcolonmathbbR^2tomathbbR$ with $f(x,y):=g(x)+g(y)$ has the properties to be continuous and to have no directional derivatives at all.
Edit: I'm rather sure that my claim is true, but it seems to be not so easy to proof it, as the fact that two limits do not exist does not always imply that the limit of the sum also does not exist.
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
add a comment |Â
up vote
0
down vote
The question in the title asks, at least in my understandig, also whether there are functions for which no directional derivatives exist at all.
In dead such (continuous) functions exist. Take any continuous nowhere differentiable function $gcolonmathbbRtomathbbR$. Then $fcolonmathbbR^2tomathbbR$ with $f(x,y):=g(x)+g(y)$ has the properties to be continuous and to have no directional derivatives at all.
Edit: I'm rather sure that my claim is true, but it seems to be not so easy to proof it, as the fact that two limits do not exist does not always imply that the limit of the sum also does not exist.
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The question in the title asks, at least in my understandig, also whether there are functions for which no directional derivatives exist at all.
In dead such (continuous) functions exist. Take any continuous nowhere differentiable function $gcolonmathbbRtomathbbR$. Then $fcolonmathbbR^2tomathbbR$ with $f(x,y):=g(x)+g(y)$ has the properties to be continuous and to have no directional derivatives at all.
Edit: I'm rather sure that my claim is true, but it seems to be not so easy to proof it, as the fact that two limits do not exist does not always imply that the limit of the sum also does not exist.
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
The question in the title asks, at least in my understandig, also whether there are functions for which no directional derivatives exist at all.
In dead such (continuous) functions exist. Take any continuous nowhere differentiable function $gcolonmathbbRtomathbbR$. Then $fcolonmathbbR^2tomathbbR$ with $f(x,y):=g(x)+g(y)$ has the properties to be continuous and to have no directional derivatives at all.
Edit: I'm rather sure that my claim is true, but it seems to be not so easy to proof it, as the fact that two limits do not exist does not always imply that the limit of the sum also does not exist.
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
edited Jul 24 at 9:22
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
answered Jul 24 at 7:21
Jens Schwaiger
1,112116
1,112116
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861031%2fcan-the-non-differentiability-of-a-function-frn-to-r-always-be-proved-by-us%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
The question in the title is completely different from the question in the post
– Jack M
Jul 24 at 7:40
Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_v+wf = D_vf + D_wf$ for all $v,w$ (note that $D_lambda vf = lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature.
– Paul Frost
Jul 24 at 8:03