Pointwise convergence when $f_n$ is defined as sum of other functions
I have this problem:
Let $f: [0,1] longrightarrow mathbb{R}$ be a continuous function. Now, define:
$$ f_n (x) = sum_{i=0}^{n} {n choose i}x^{i}(1-x)^{n-i} f( frac{i}{n})$$
Show that $f_n$ converges pointwise to f.
I worked (succesfully) the case in which $f$ is the identity map. However, I cant't handle the general case. Could someone give me pointers in how to deal with this kind of problem ? (when the functions in the sequence not only depends on $n$ and $x$, but also in some other index. Thanks in advance!
PD: This problem was left by my teacher to prepare the third exam in my class of probability theory.(the last thing we saw was the Law of large numbers). So maybe there is a natural way of attacking this problem using only probability theory but (honestly) I don't see how.
calculus real-analysis probability-theory
asked Nov 21 '18 at 3:43
JuanuPE
82
add a comment |
I have this problem:
Let $f: [0,1] longrightarrow mathbb{R}$ be a continuous function. Now, define:
$$ f_n (x) = sum_{i=0}^{n} {n choose i}x^{i}(1-x)^{n-i} f( frac{i}{n})$$
Show that $f_n$ converges pointwise to f.
I worked (succesfully) the case in which $f$ is the identity map. However, I cant't handle the general case. Could someone give me pointers in how to deal with this kind of problem ? (when the functions in the sequence not only depends on $n$ and $x$, but also in some other index. Thanks in advance!
PD: This problem was left by my teacher to prepare the third exam in my class of probability theory.(the last thing we saw was the Law of large numbers). So maybe there is a natural way of attacking this problem using only probability theory but (honestly) I don't see how.
calculus real-analysis probability-theory
asked Nov 21 '18 at 3:43
JuanuPE
82
1
en.wikipedia.org/wiki/…
– Lord Shark the Unknown
Nov 21 '18 at 3:45
I see, so it's a famous problem. Thanks a lot!
– JuanuPE
Nov 21 '18 at 3:53
add a comment |
I have this problem:
Let $f: [0,1] longrightarrow mathbb{R}$ be a continuous function. Now, define:
$$ f_n (x) = sum_{i=0}^{n} {n choose i}x^{i}(1-x)^{n-i} f( frac{i}{n})$$
Show that $f_n$ converges pointwise to f.
I worked (succesfully) the case in which $f$ is the identity map. However, I cant't handle the general case. Could someone give me pointers in how to deal with this kind of problem ? (when the functions in the sequence not only depends on $n$ and $x$, but also in some other index. Thanks in advance!
PD: This problem was left by my teacher to prepare the third exam in my class of probability theory.(the last thing we saw was the Law of large numbers). So maybe there is a natural way of attacking this problem using only probability theory but (honestly) I don't see how.
calculus real-analysis probability-theory
asked Nov 21 '18 at 3:43
JuanuPE
82
I have this problem:
Let $f: [0,1] longrightarrow mathbb{R}$ be a continuous function. Now, define:
$$ f_n (x) = sum_{i=0}^{n} {n choose i}x^{i}(1-x)^{n-i} f( frac{i}{n})$$
Show that $f_n$ converges pointwise to f.
I worked (succesfully) the case in which $f$ is the identity map. However, I cant't handle the general case. Could someone give me pointers in how to deal with this kind of problem ? (when the functions in the sequence not only depends on $n$ and $x$, but also in some other index. Thanks in advance!
PD: This problem was left by my teacher to prepare the third exam in my class of probability theory.(the last thing we saw was the Law of large numbers). So maybe there is a natural way of attacking this problem using only probability theory but (honestly) I don't see how.
calculus real-analysis probability-theory
calculus real-analysis probability-theory
asked Nov 21 '18 at 3:43
JuanuPE
82
asked Nov 21 '18 at 3:43
JuanuPE
82
asked Nov 21 '18 at 3:43
JuanuPE
82
asked Nov 21 '18 at 3:43
JuanuPE
82
asked Nov 21 '18 at 3:43
JuanuPE
82
82
1
en.wikipedia.org/wiki/…
– Lord Shark the Unknown
Nov 21 '18 at 3:45
I see, so it's a famous problem. Thanks a lot!
– JuanuPE
Nov 21 '18 at 3:53
add a comment |
1
en.wikipedia.org/wiki/…
– Lord Shark the Unknown
Nov 21 '18 at 3:45
I see, so it's a famous problem. Thanks a lot!
– JuanuPE
Nov 21 '18 at 3:53
1
1
en.wikipedia.org/wiki/…
– Lord Shark the Unknown
Nov 21 '18 at 3:45
en.wikipedia.org/wiki/…
– Lord Shark the Unknown
Nov 21 '18 at 3:45
I see, so it's a famous problem. Thanks a lot!
– JuanuPE
Nov 21 '18 at 3:53
I see, so it's a famous problem. Thanks a lot!
– JuanuPE
Nov 21 '18 at 3:53
add a comment |
1 Answer
1
active
oldest
votes
We could also take i.i.d. random variables $(X_i)_{i in mathbb{N}}$ with probability $P(X_i=0) = (1-x)$ and $P(X_i=1) = x$. Let now $S_n = sum_{k=1}^n$ X_i. By the strong law of large numbers we have $$frac{1}{n} S_n rightarrow x$$
$mathbb{P}$-almost sure. This also implies convergence in probablity. Since $f$ is bounded on $[0,1]$, we see that
$$sum_{k=0}^n binom{n}{k} x^k (1-x)^{n-k} f(k/n) = mathbb{E}[f(S_n/n)] rightarrow mathbb{E}[f(x)] = f(x).$$
This gives pointwise convergence. In the proof of the Wikipedia article - see here - is shown that this convergence is uniformly in x. (This proves the famous Weierstraß theorem on the approximation of functions $C[0,1]$ by polynomials.)
answered Nov 21 '18 at 7:51
p4sch
4,760217
add a comment |
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%2f3007230%2fpointwise-convergence-when-f-n-is-defined-as-sum-of-other-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We could also take i.i.d. random variables $(X_i)_{i in mathbb{N}}$ with probability $P(X_i=0) = (1-x)$ and $P(X_i=1) = x$. Let now $S_n = sum_{k=1}^n$ X_i. By the strong law of large numbers we have $$frac{1}{n} S_n rightarrow x$$
$mathbb{P}$-almost sure. This also implies convergence in probablity. Since $f$ is bounded on $[0,1]$, we see that
$$sum_{k=0}^n binom{n}{k} x^k (1-x)^{n-k} f(k/n) = mathbb{E}[f(S_n/n)] rightarrow mathbb{E}[f(x)] = f(x).$$
This gives pointwise convergence. In the proof of the Wikipedia article - see here - is shown that this convergence is uniformly in x. (This proves the famous Weierstraß theorem on the approximation of functions $C[0,1]$ by polynomials.)
answered Nov 21 '18 at 7:51
p4sch
4,760217
add a comment |
We could also take i.i.d. random variables $(X_i)_{i in mathbb{N}}$ with probability $P(X_i=0) = (1-x)$ and $P(X_i=1) = x$. Let now $S_n = sum_{k=1}^n$ X_i. By the strong law of large numbers we have $$frac{1}{n} S_n rightarrow x$$
$mathbb{P}$-almost sure. This also implies convergence in probablity. Since $f$ is bounded on $[0,1]$, we see that
$$sum_{k=0}^n binom{n}{k} x^k (1-x)^{n-k} f(k/n) = mathbb{E}[f(S_n/n)] rightarrow mathbb{E}[f(x)] = f(x).$$
This gives pointwise convergence. In the proof of the Wikipedia article - see here - is shown that this convergence is uniformly in x. (This proves the famous Weierstraß theorem on the approximation of functions $C[0,1]$ by polynomials.)
answered Nov 21 '18 at 7:51
p4sch
4,760217
add a comment |
We could also take i.i.d. random variables $(X_i)_{i in mathbb{N}}$ with probability $P(X_i=0) = (1-x)$ and $P(X_i=1) = x$. Let now $S_n = sum_{k=1}^n$ X_i. By the strong law of large numbers we have $$frac{1}{n} S_n rightarrow x$$
$mathbb{P}$-almost sure. This also implies convergence in probablity. Since $f$ is bounded on $[0,1]$, we see that
$$sum_{k=0}^n binom{n}{k} x^k (1-x)^{n-k} f(k/n) = mathbb{E}[f(S_n/n)] rightarrow mathbb{E}[f(x)] = f(x).$$
This gives pointwise convergence. In the proof of the Wikipedia article - see here - is shown that this convergence is uniformly in x. (This proves the famous Weierstraß theorem on the approximation of functions $C[0,1]$ by polynomials.)
answered Nov 21 '18 at 7:51
p4sch
4,760217
We could also take i.i.d. random variables $(X_i)_{i in mathbb{N}}$ with probability $P(X_i=0) = (1-x)$ and $P(X_i=1) = x$. Let now $S_n = sum_{k=1}^n$ X_i. By the strong law of large numbers we have $$frac{1}{n} S_n rightarrow x$$
$mathbb{P}$-almost sure. This also implies convergence in probablity. Since $f$ is bounded on $[0,1]$, we see that
$$sum_{k=0}^n binom{n}{k} x^k (1-x)^{n-k} f(k/n) = mathbb{E}[f(S_n/n)] rightarrow mathbb{E}[f(x)] = f(x).$$
This gives pointwise convergence. In the proof of the Wikipedia article - see here - is shown that this convergence is uniformly in x. (This proves the famous Weierstraß theorem on the approximation of functions $C[0,1]$ by polynomials.)
answered Nov 21 '18 at 7:51
p4sch
4,760217
answered Nov 21 '18 at 7:51
p4sch
4,760217
answered Nov 21 '18 at 7:51
p4sch
4,760217
answered Nov 21 '18 at 7:51
p4sch
4,760217
4,760217
add a comment |
add a comment |
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.
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%2f3007230%2fpointwise-convergence-when-f-n-is-defined-as-sum-of-other-functions%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
1
en.wikipedia.org/wiki/…
– Lord Shark the Unknown
Nov 21 '18 at 3:45
I see, so it's a famous problem. Thanks a lot!
– JuanuPE
Nov 21 '18 at 3:53