Product of discontinuous and smooth function
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Let $f$ be a smooth function (i.e. $C^infty(mathbb{R}^n)$), $g$ discontinuous in some $x_0in mathbb{R}^n$ and smooth everywhere else, for example $g(x)=Theta(x)$ or $g(x)=frac{1}{x}$. Furthermore, $fgin C^infty(mathbb{R}^n)$.
I want to show, that this already implies, that all partial derivatives of $f$ vanish at $x_0$.
Intuitively, this seems right as $f$ has to smooth out $g$ at $x_0$. One can easliy show, that $f(x_0)=0$: assume the contrary, then $g=frac{fg}{f}$ would be smooth at $x_0$, which contradicts the assumption. However, this trick does not work for derivatives. I also tried fooling around with some limits, didn't get to anything substantial so far.
I still think, that this has to be quite elementary. Does anyone have an idea?
real-analysis continuity
asked Dec 4 '14 at 1:26
Daniel
601413
add a comment |
up vote
1
down vote
favorite
Let $f$ be a smooth function (i.e. $C^infty(mathbb{R}^n)$), $g$ discontinuous in some $x_0in mathbb{R}^n$ and smooth everywhere else, for example $g(x)=Theta(x)$ or $g(x)=frac{1}{x}$. Furthermore, $fgin C^infty(mathbb{R}^n)$.
I want to show, that this already implies, that all partial derivatives of $f$ vanish at $x_0$.
Intuitively, this seems right as $f$ has to smooth out $g$ at $x_0$. One can easliy show, that $f(x_0)=0$: assume the contrary, then $g=frac{fg}{f}$ would be smooth at $x_0$, which contradicts the assumption. However, this trick does not work for derivatives. I also tried fooling around with some limits, didn't get to anything substantial so far.
I still think, that this has to be quite elementary. Does anyone have an idea?
real-analysis continuity
asked Dec 4 '14 at 1:26
Daniel
601413
2
With $g(x) = {1 over x} $ and $f(x) = x$ we have $fg=1$, but the partials of $f$ do not vanish.
– copper.hat
Dec 4 '14 at 1:48
Ah, crap, that seems to be true. Guess I'll need to search for more conditions then. Thank you!
– Daniel
Dec 4 '14 at 14:16
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f$ be a smooth function (i.e. $C^infty(mathbb{R}^n)$), $g$ discontinuous in some $x_0in mathbb{R}^n$ and smooth everywhere else, for example $g(x)=Theta(x)$ or $g(x)=frac{1}{x}$. Furthermore, $fgin C^infty(mathbb{R}^n)$.
I want to show, that this already implies, that all partial derivatives of $f$ vanish at $x_0$.
Intuitively, this seems right as $f$ has to smooth out $g$ at $x_0$. One can easliy show, that $f(x_0)=0$: assume the contrary, then $g=frac{fg}{f}$ would be smooth at $x_0$, which contradicts the assumption. However, this trick does not work for derivatives. I also tried fooling around with some limits, didn't get to anything substantial so far.
I still think, that this has to be quite elementary. Does anyone have an idea?
real-analysis continuity
asked Dec 4 '14 at 1:26
Daniel
601413
Let $f$ be a smooth function (i.e. $C^infty(mathbb{R}^n)$), $g$ discontinuous in some $x_0in mathbb{R}^n$ and smooth everywhere else, for example $g(x)=Theta(x)$ or $g(x)=frac{1}{x}$. Furthermore, $fgin C^infty(mathbb{R}^n)$.
I want to show, that this already implies, that all partial derivatives of $f$ vanish at $x_0$.
Intuitively, this seems right as $f$ has to smooth out $g$ at $x_0$. One can easliy show, that $f(x_0)=0$: assume the contrary, then $g=frac{fg}{f}$ would be smooth at $x_0$, which contradicts the assumption. However, this trick does not work for derivatives. I also tried fooling around with some limits, didn't get to anything substantial so far.
I still think, that this has to be quite elementary. Does anyone have an idea?
real-analysis continuity
real-analysis continuity
asked Dec 4 '14 at 1:26
Daniel
601413
asked Dec 4 '14 at 1:26
Daniel
601413
asked Dec 4 '14 at 1:26
Daniel
601413
asked Dec 4 '14 at 1:26
Daniel
601413
asked Dec 4 '14 at 1:26
Daniel
601413
601413
2
With $g(x) = {1 over x} $ and $f(x) = x$ we have $fg=1$, but the partials of $f$ do not vanish.
– copper.hat
Dec 4 '14 at 1:48
Ah, crap, that seems to be true. Guess I'll need to search for more conditions then. Thank you!
– Daniel
Dec 4 '14 at 14:16
add a comment |
2
With $g(x) = {1 over x} $ and $f(x) = x$ we have $fg=1$, but the partials of $f$ do not vanish.
– copper.hat
Dec 4 '14 at 1:48
Ah, crap, that seems to be true. Guess I'll need to search for more conditions then. Thank you!
– Daniel
Dec 4 '14 at 14:16
2
2
With $g(x) = {1 over x} $ and $f(x) = x$ we have $fg=1$, but the partials of $f$ do not vanish.
– copper.hat
Dec 4 '14 at 1:48
With $g(x) = {1 over x} $ and $f(x) = x$ we have $fg=1$, but the partials of $f$ do not vanish.
– copper.hat
Dec 4 '14 at 1:48
Ah, crap, that seems to be true. Guess I'll need to search for more conditions then. Thank you!
– Daniel
Dec 4 '14 at 14:16
Ah, crap, that seems to be true. Guess I'll need to search for more conditions then. Thank you!
– Daniel
Dec 4 '14 at 14:16
add a comment |
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2
With $g(x) = {1 over x} $ and $f(x) = x$ we have $fg=1$, but the partials of $f$ do not vanish.
– copper.hat
Dec 4 '14 at 1:48
Ah, crap, that seems to be true. Guess I'll need to search for more conditions then. Thank you!
– Daniel
Dec 4 '14 at 14:16