If $f(x+a)-f(x)$ is differentiable for each $a$, then $f$ is differentiable
$begingroup$
$ f in C(mathbf{R})$, if for each real $a$, $f(x+a)-f(x)$ is differentiable, then $f$ is differentiable.
It seems hard to convert difference to the original function
real-analysis derivatives
edited Dec 24 '18 at 8:32
bof
52.6k559121
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
$endgroup$
add a comment |
$begingroup$
$ f in C(mathbf{R})$, if for each real $a$, $f(x+a)-f(x)$ is differentiable, then $f$ is differentiable.
It seems hard to convert difference to the original function
real-analysis derivatives
edited Dec 24 '18 at 8:32
bof
52.6k559121
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
$endgroup$
$begingroup$
Is the hypothesis that there exists an $a$ so that $f(x+a)-f(x)$ is differentiable or is it that for each $a$ the function $xmapsto f(x+a)-f(x)$ is differentiable?
$endgroup$
– kimchi lover
Dec 24 '18 at 3:47
$begingroup$
There's an "if" missing at some point before the "then", and it matters for the meaning where you put it. Please edit the question.
$endgroup$
– Henning Makholm
Dec 24 '18 at 3:47
1
$begingroup$
@kimchilover It has to be for all $a$, otherwise any periodic nondifferentiable function is a counterexample...
$endgroup$
– N. S.
Dec 24 '18 at 3:48
add a comment |
$begingroup$
$ f in C(mathbf{R})$, if for each real $a$, $f(x+a)-f(x)$ is differentiable, then $f$ is differentiable.
It seems hard to convert difference to the original function
real-analysis derivatives
edited Dec 24 '18 at 8:32
bof
52.6k559121
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
$endgroup$
$ f in C(mathbf{R})$, if for each real $a$, $f(x+a)-f(x)$ is differentiable, then $f$ is differentiable.
It seems hard to convert difference to the original function
real-analysis derivatives
real-analysis derivatives
edited Dec 24 '18 at 8:32
bof
52.6k559121
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
edited Dec 24 '18 at 8:32
bof
52.6k559121
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
edited Dec 24 '18 at 8:32
bof
52.6k559121
edited Dec 24 '18 at 8:32
bof
52.6k559121
edited Dec 24 '18 at 8:32
bof
52.6k559121
52.6k559121
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
asked Dec 24 '18 at 3:27
Rhove LuoRhove Luo
395
395
$begingroup$
Is the hypothesis that there exists an $a$ so that $f(x+a)-f(x)$ is differentiable or is it that for each $a$ the function $xmapsto f(x+a)-f(x)$ is differentiable?
$endgroup$
– kimchi lover
Dec 24 '18 at 3:47
$begingroup$
There's an "if" missing at some point before the "then", and it matters for the meaning where you put it. Please edit the question.
$endgroup$
– Henning Makholm
Dec 24 '18 at 3:47
1
$begingroup$
@kimchilover It has to be for all $a$, otherwise any periodic nondifferentiable function is a counterexample...
$endgroup$
– N. S.
Dec 24 '18 at 3:48
add a comment |
$begingroup$
Is the hypothesis that there exists an $a$ so that $f(x+a)-f(x)$ is differentiable or is it that for each $a$ the function $xmapsto f(x+a)-f(x)$ is differentiable?
$endgroup$
– kimchi lover
Dec 24 '18 at 3:47
$begingroup$
There's an "if" missing at some point before the "then", and it matters for the meaning where you put it. Please edit the question.
$endgroup$
– Henning Makholm
Dec 24 '18 at 3:47
1
$begingroup$
@kimchilover It has to be for all $a$, otherwise any periodic nondifferentiable function is a counterexample...
$endgroup$
– N. S.
Dec 24 '18 at 3:48
$begingroup$
Is the hypothesis that there exists an $a$ so that $f(x+a)-f(x)$ is differentiable or is it that for each $a$ the function $xmapsto f(x+a)-f(x)$ is differentiable?
$endgroup$
– kimchi lover
Dec 24 '18 at 3:47
$begingroup$
Is the hypothesis that there exists an $a$ so that $f(x+a)-f(x)$ is differentiable or is it that for each $a$ the function $xmapsto f(x+a)-f(x)$ is differentiable?
$endgroup$
– kimchi lover
Dec 24 '18 at 3:47
$begingroup$
There's an "if" missing at some point before the "then", and it matters for the meaning where you put it. Please edit the question.
$endgroup$
– Henning Makholm
Dec 24 '18 at 3:47
$begingroup$
There's an "if" missing at some point before the "then", and it matters for the meaning where you put it. Please edit the question.
$endgroup$
– Henning Makholm
Dec 24 '18 at 3:47
1
1
$begingroup$
@kimchilover It has to be for all $a$, otherwise any periodic nondifferentiable function is a counterexample...
$endgroup$
– N. S.
Dec 24 '18 at 3:48
$begingroup$
@kimchilover It has to be for all $a$, otherwise any periodic nondifferentiable function is a counterexample...
$endgroup$
– N. S.
Dec 24 '18 at 3:48
add a comment |
1 Answer
1
active
oldest
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$begingroup$
This and many other results of this kind are proved in the paper
Bruijn, de, N. G. (1951). Functions whose differences belong to a given class. Nieuw Archief voor Wiskunde, serie 2, 23, 194-218.
The assumption that $f(x)$ is continuous can be weakened, e.g., it's enough to assume that $f(x)$ is bounded on a set of positive measure. Some condition on $f(x)$ is needed since, if $f(x)$ is a discontinuous additive function, then $xmapsto f(x+a)-f(x)$ is the constant function $xmapsto f(a)$.
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
$endgroup$
add a comment |
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$begingroup$
This and many other results of this kind are proved in the paper
Bruijn, de, N. G. (1951). Functions whose differences belong to a given class. Nieuw Archief voor Wiskunde, serie 2, 23, 194-218.
The assumption that $f(x)$ is continuous can be weakened, e.g., it's enough to assume that $f(x)$ is bounded on a set of positive measure. Some condition on $f(x)$ is needed since, if $f(x)$ is a discontinuous additive function, then $xmapsto f(x+a)-f(x)$ is the constant function $xmapsto f(a)$.
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
This and many other results of this kind are proved in the paper
Bruijn, de, N. G. (1951). Functions whose differences belong to a given class. Nieuw Archief voor Wiskunde, serie 2, 23, 194-218.
The assumption that $f(x)$ is continuous can be weakened, e.g., it's enough to assume that $f(x)$ is bounded on a set of positive measure. Some condition on $f(x)$ is needed since, if $f(x)$ is a discontinuous additive function, then $xmapsto f(x+a)-f(x)$ is the constant function $xmapsto f(a)$.
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
This and many other results of this kind are proved in the paper
Bruijn, de, N. G. (1951). Functions whose differences belong to a given class. Nieuw Archief voor Wiskunde, serie 2, 23, 194-218.
The assumption that $f(x)$ is continuous can be weakened, e.g., it's enough to assume that $f(x)$ is bounded on a set of positive measure. Some condition on $f(x)$ is needed since, if $f(x)$ is a discontinuous additive function, then $xmapsto f(x+a)-f(x)$ is the constant function $xmapsto f(a)$.
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
$endgroup$
This and many other results of this kind are proved in the paper
Bruijn, de, N. G. (1951). Functions whose differences belong to a given class. Nieuw Archief voor Wiskunde, serie 2, 23, 194-218.
The assumption that $f(x)$ is continuous can be weakened, e.g., it's enough to assume that $f(x)$ is bounded on a set of positive measure. Some condition on $f(x)$ is needed since, if $f(x)$ is a discontinuous additive function, then $xmapsto f(x+a)-f(x)$ is the constant function $xmapsto f(a)$.
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
edited Dec 24 '18 at 5:20
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
answered Dec 24 '18 at 4:54
bofbof
52.6k559121
52.6k559121
add a comment |
add a comment |
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$begingroup$
Is the hypothesis that there exists an $a$ so that $f(x+a)-f(x)$ is differentiable or is it that for each $a$ the function $xmapsto f(x+a)-f(x)$ is differentiable?
$endgroup$
– kimchi lover
Dec 24 '18 at 3:47
$begingroup$
There's an "if" missing at some point before the "then", and it matters for the meaning where you put it. Please edit the question.
$endgroup$
– Henning Makholm
Dec 24 '18 at 3:47
1
$begingroup$
@kimchilover It has to be for all $a$, otherwise any periodic nondifferentiable function is a counterexample...
$endgroup$
– N. S.
Dec 24 '18 at 3:48