Mixing
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
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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
4,770217
add a comment |
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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
4,770217
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
4,770217
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
4,770217
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
4,770217
answered Nov 21 '18 at 7:51
p4sch
4,770217
answered Nov 21 '18 at 7:51
p4sch
4,770217
answered Nov 21 '18 at 7:51
p4sch
4,770217
4,770217
add a comment |
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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