Mixing
show that $fp(frac{1}{x^2})$ defines a distribution?
$begingroup$
let $phi$ be a test function, let $epsilon > 0$, then :
$$langle fp(frac{1}{x^2}),phirangle = lim_{epsilon to 0} [int_{|x| geq epsilon} frac{phi(x)}{x^2}dx - 2frac{phi(0)}{epsilon}]$$
show that this defines a distribution
my attempt :
let $K = [-m,m]$ such that $suppphi subset K$
let's first rewrite the expression without the limit :
$$int_{|x| geq epsilon} frac{phi(x)}{x^2} dx - 2frac{phi(0)}{epsilon} = int_{|x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx$$
let's apply a taylor expansion with integral remainder to $phi$
$$phi(x) = phi(0)+xphi'(0) +x^2int_0^1phi''(xt)(1-t)dt = phi(0)+xphi'(0) +x^2psi(x)$$
we also have $|psi(x)| leq sup_{K} |phi''|$
so putting all these back into the equation we get :
$$int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi'(0) }{x} dx + int_{mgeq |x| geq epsilon} psi(x) dx$$
first term is zero because we're integrating an odd function over a symetric interval.
and by the dominated convergence theorem we get :
$$langle fp(frac{1}{x^2}),phirangle = int_{mgeq |x|} psi(x) dx $$
and
$$|langle fp(frac{1}{x^2}),phirangle| leq 2m sup_K|phi''|$$
so it defines a distribution of order $leq 2$
linearity is obvious so I think I'm done.
is my work all right ?
proof-verification distribution-theory
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
$endgroup$
add a comment |
$begingroup$
let $phi$ be a test function, let $epsilon > 0$, then :
$$langle fp(frac{1}{x^2}),phirangle = lim_{epsilon to 0} [int_{|x| geq epsilon} frac{phi(x)}{x^2}dx - 2frac{phi(0)}{epsilon}]$$
show that this defines a distribution
my attempt :
let $K = [-m,m]$ such that $suppphi subset K$
let's first rewrite the expression without the limit :
$$int_{|x| geq epsilon} frac{phi(x)}{x^2} dx - 2frac{phi(0)}{epsilon} = int_{|x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx$$
let's apply a taylor expansion with integral remainder to $phi$
$$phi(x) = phi(0)+xphi'(0) +x^2int_0^1phi''(xt)(1-t)dt = phi(0)+xphi'(0) +x^2psi(x)$$
we also have $|psi(x)| leq sup_{K} |phi''|$
so putting all these back into the equation we get :
$$int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi'(0) }{x} dx + int_{mgeq |x| geq epsilon} psi(x) dx$$
first term is zero because we're integrating an odd function over a symetric interval.
and by the dominated convergence theorem we get :
$$langle fp(frac{1}{x^2}),phirangle = int_{mgeq |x|} psi(x) dx $$
and
$$|langle fp(frac{1}{x^2}),phirangle| leq 2m sup_K|phi''|$$
so it defines a distribution of order $leq 2$
linearity is obvious so I think I'm done.
is my work all right ?
proof-verification distribution-theory
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
$endgroup$
add a comment |
$begingroup$
let $phi$ be a test function, let $epsilon > 0$, then :
$$langle fp(frac{1}{x^2}),phirangle = lim_{epsilon to 0} [int_{|x| geq epsilon} frac{phi(x)}{x^2}dx - 2frac{phi(0)}{epsilon}]$$
show that this defines a distribution
my attempt :
let $K = [-m,m]$ such that $suppphi subset K$
let's first rewrite the expression without the limit :
$$int_{|x| geq epsilon} frac{phi(x)}{x^2} dx - 2frac{phi(0)}{epsilon} = int_{|x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx$$
let's apply a taylor expansion with integral remainder to $phi$
$$phi(x) = phi(0)+xphi'(0) +x^2int_0^1phi''(xt)(1-t)dt = phi(0)+xphi'(0) +x^2psi(x)$$
we also have $|psi(x)| leq sup_{K} |phi''|$
so putting all these back into the equation we get :
$$int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi'(0) }{x} dx + int_{mgeq |x| geq epsilon} psi(x) dx$$
first term is zero because we're integrating an odd function over a symetric interval.
and by the dominated convergence theorem we get :
$$langle fp(frac{1}{x^2}),phirangle = int_{mgeq |x|} psi(x) dx $$
and
$$|langle fp(frac{1}{x^2}),phirangle| leq 2m sup_K|phi''|$$
so it defines a distribution of order $leq 2$
linearity is obvious so I think I'm done.
is my work all right ?
proof-verification distribution-theory
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
$endgroup$
let $phi$ be a test function, let $epsilon > 0$, then :
$$langle fp(frac{1}{x^2}),phirangle = lim_{epsilon to 0} [int_{|x| geq epsilon} frac{phi(x)}{x^2}dx - 2frac{phi(0)}{epsilon}]$$
show that this defines a distribution
my attempt :
let $K = [-m,m]$ such that $suppphi subset K$
let's first rewrite the expression without the limit :
$$int_{|x| geq epsilon} frac{phi(x)}{x^2} dx - 2frac{phi(0)}{epsilon} = int_{|x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx$$
let's apply a taylor expansion with integral remainder to $phi$
$$phi(x) = phi(0)+xphi'(0) +x^2int_0^1phi''(xt)(1-t)dt = phi(0)+xphi'(0) +x^2psi(x)$$
we also have $|psi(x)| leq sup_{K} |phi''|$
so putting all these back into the equation we get :
$$int_{mgeq |x| geq epsilon} frac{phi(x) - phi(0)}{x^2} dx = int_{mgeq |x| geq epsilon} frac{phi'(0) }{x} dx + int_{mgeq |x| geq epsilon} psi(x) dx$$
first term is zero because we're integrating an odd function over a symetric interval.
and by the dominated convergence theorem we get :
$$langle fp(frac{1}{x^2}),phirangle = int_{mgeq |x|} psi(x) dx $$
and
$$|langle fp(frac{1}{x^2}),phirangle| leq 2m sup_K|phi''|$$
so it defines a distribution of order $leq 2$
linearity is obvious so I think I'm done.
is my work all right ?
proof-verification distribution-theory
proof-verification distribution-theory
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
asked Jan 28 at 16:44
rapidracimrapidracim
1,7291419
1,7291419
add a comment |
add a comment |
0
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',
autoActivateHeartbeat: false,
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091099%2fshow-that-fp-frac1x2-defines-a-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091099%2fshow-that-fp-frac1x2-defines-a-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
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
