Prove that a function is smooth if it is smooth in almost all directions
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Question
So suppose we have a function $f:mathbb R^2to mathbb R$ for which it is given that $xmapsto f(x,g(x))$ is smooth (i.e., $C^infty$) for all smooth functions $g:mathbb Rtomathbb R$. Can we prove that $f$ is smooth as well?
I don't know whether this statement is true and honestly I wouldn't be surprised either way.
What I've tried already
Fix a point $(x_0,y_0)$. Intuitively, by taking $g(x) = lambda x$ with $lambdainmathbb R$ we see that $f$ should be at least differentiable along all directions $(1,lambda)$ at $(x_0,y_0)$. This follows for instance by considering the curve $tmapsto (x_0+t, y_0+lambda t)$. Thus the only direction that is non-trivial is the vertical direction $(0,1)$. If we can show that $f$ is also differentiable in that direction then I'm confident that it will be possible to show that $f$ is differentiable. But how can we show whether $f$ is differentiable along $(0,1)$? We cannot do it directly from the fact that $f(x,g(x))$ is smooth, but perhaps we can use a limiting argument, letting the slope of the curve $(t,g(t))$ tend to infinity?
When we know that $f$ is differentiable, it will probably be possible using an inductive argument to prove that $f$ is smooth (i.e., $C^infty$).
Any help is appreciated.
EDIT. If found a closely related result, namely Boman's theorem, which says basically says that $f$ is smooth if and only if $fcircgamma$ is smooth for all smooth curves $gamma:mathbb Rtomathbb R^2$. I feel like the statement of my question should probably be reducible to this theorem. The only difficulty is that we don't necessarily know if our $f$ is differentiable along vertical curves, but perhaps this follows in some way.
real-analysis
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
$endgroup$
add a comment |
$begingroup$
Question
So suppose we have a function $f:mathbb R^2to mathbb R$ for which it is given that $xmapsto f(x,g(x))$ is smooth (i.e., $C^infty$) for all smooth functions $g:mathbb Rtomathbb R$. Can we prove that $f$ is smooth as well?
I don't know whether this statement is true and honestly I wouldn't be surprised either way.
What I've tried already
Fix a point $(x_0,y_0)$. Intuitively, by taking $g(x) = lambda x$ with $lambdainmathbb R$ we see that $f$ should be at least differentiable along all directions $(1,lambda)$ at $(x_0,y_0)$. This follows for instance by considering the curve $tmapsto (x_0+t, y_0+lambda t)$. Thus the only direction that is non-trivial is the vertical direction $(0,1)$. If we can show that $f$ is also differentiable in that direction then I'm confident that it will be possible to show that $f$ is differentiable. But how can we show whether $f$ is differentiable along $(0,1)$? We cannot do it directly from the fact that $f(x,g(x))$ is smooth, but perhaps we can use a limiting argument, letting the slope of the curve $(t,g(t))$ tend to infinity?
When we know that $f$ is differentiable, it will probably be possible using an inductive argument to prove that $f$ is smooth (i.e., $C^infty$).
Any help is appreciated.
EDIT. If found a closely related result, namely Boman's theorem, which says basically says that $f$ is smooth if and only if $fcircgamma$ is smooth for all smooth curves $gamma:mathbb Rtomathbb R^2$. I feel like the statement of my question should probably be reducible to this theorem. The only difficulty is that we don't necessarily know if our $f$ is differentiable along vertical curves, but perhaps this follows in some way.
real-analysis
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
$endgroup$
add a comment |
$begingroup$
Question
So suppose we have a function $f:mathbb R^2to mathbb R$ for which it is given that $xmapsto f(x,g(x))$ is smooth (i.e., $C^infty$) for all smooth functions $g:mathbb Rtomathbb R$. Can we prove that $f$ is smooth as well?
I don't know whether this statement is true and honestly I wouldn't be surprised either way.
What I've tried already
Fix a point $(x_0,y_0)$. Intuitively, by taking $g(x) = lambda x$ with $lambdainmathbb R$ we see that $f$ should be at least differentiable along all directions $(1,lambda)$ at $(x_0,y_0)$. This follows for instance by considering the curve $tmapsto (x_0+t, y_0+lambda t)$. Thus the only direction that is non-trivial is the vertical direction $(0,1)$. If we can show that $f$ is also differentiable in that direction then I'm confident that it will be possible to show that $f$ is differentiable. But how can we show whether $f$ is differentiable along $(0,1)$? We cannot do it directly from the fact that $f(x,g(x))$ is smooth, but perhaps we can use a limiting argument, letting the slope of the curve $(t,g(t))$ tend to infinity?
When we know that $f$ is differentiable, it will probably be possible using an inductive argument to prove that $f$ is smooth (i.e., $C^infty$).
Any help is appreciated.
EDIT. If found a closely related result, namely Boman's theorem, which says basically says that $f$ is smooth if and only if $fcircgamma$ is smooth for all smooth curves $gamma:mathbb Rtomathbb R^2$. I feel like the statement of my question should probably be reducible to this theorem. The only difficulty is that we don't necessarily know if our $f$ is differentiable along vertical curves, but perhaps this follows in some way.
real-analysis
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
$endgroup$
Question
So suppose we have a function $f:mathbb R^2to mathbb R$ for which it is given that $xmapsto f(x,g(x))$ is smooth (i.e., $C^infty$) for all smooth functions $g:mathbb Rtomathbb R$. Can we prove that $f$ is smooth as well?
I don't know whether this statement is true and honestly I wouldn't be surprised either way.
What I've tried already
Fix a point $(x_0,y_0)$. Intuitively, by taking $g(x) = lambda x$ with $lambdainmathbb R$ we see that $f$ should be at least differentiable along all directions $(1,lambda)$ at $(x_0,y_0)$. This follows for instance by considering the curve $tmapsto (x_0+t, y_0+lambda t)$. Thus the only direction that is non-trivial is the vertical direction $(0,1)$. If we can show that $f$ is also differentiable in that direction then I'm confident that it will be possible to show that $f$ is differentiable. But how can we show whether $f$ is differentiable along $(0,1)$? We cannot do it directly from the fact that $f(x,g(x))$ is smooth, but perhaps we can use a limiting argument, letting the slope of the curve $(t,g(t))$ tend to infinity?
When we know that $f$ is differentiable, it will probably be possible using an inductive argument to prove that $f$ is smooth (i.e., $C^infty$).
Any help is appreciated.
EDIT. If found a closely related result, namely Boman's theorem, which says basically says that $f$ is smooth if and only if $fcircgamma$ is smooth for all smooth curves $gamma:mathbb Rtomathbb R^2$. I feel like the statement of my question should probably be reducible to this theorem. The only difficulty is that we don't necessarily know if our $f$ is differentiable along vertical curves, but perhaps this follows in some way.
real-analysis
real-analysis
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
edited Jan 13 at 14:15
Sjorszini
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
asked Jan 13 at 10:28
SjorsziniSjorszini
805514
805514
add a comment |
add a comment |
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