Product of discontinuous and smooth function











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?










share|cite|improve this question


















  • 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















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?










share|cite|improve this question


















  • 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













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?










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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















active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1050772%2fproduct-of-discontinuous-and-smooth-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1050772%2fproduct-of-discontinuous-and-smooth-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Quarter-circle Tiles

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Mont Emei