Mixing
Given pointwise convergence of $f_n rightarrow f$, does $lim_{epsilon rightarrow 0} f(x+epsilon) = f(x)$...
$begingroup$
I was wondering if provided that the partial sums $f_n$ converge to $f$ matching pointwise convergence, what this would imply:
$$
lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)
$$
If that would be the case, for all $x$, wouln't this imply uniform convergence? And if not, why.
Many thanks in advance!
real-analysis convergence uniform-convergence
edited Jan 18 at 11:29
Bernard
121k740116
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
$endgroup$
add a comment |
$begingroup$
I was wondering if provided that the partial sums $f_n$ converge to $f$ matching pointwise convergence, what this would imply:
$$
lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)
$$
If that would be the case, for all $x$, wouln't this imply uniform convergence? And if not, why.
Many thanks in advance!
real-analysis convergence uniform-convergence
edited Jan 18 at 11:29
Bernard
121k740116
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
$endgroup$
$begingroup$
So $f$ is a series?
$endgroup$
– Stockfish
Jan 18 at 10:58
1
$begingroup$
This is just saying $f$ is continuous. You won't necessarily have uniform convergence. Look at $f_n(x)=chi_{(0,1/n)}$ on $[0,1]$.
$endgroup$
– David Mitra
Jan 18 at 10:59
add a comment |
$begingroup$
I was wondering if provided that the partial sums $f_n$ converge to $f$ matching pointwise convergence, what this would imply:
$$
lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)
$$
If that would be the case, for all $x$, wouln't this imply uniform convergence? And if not, why.
Many thanks in advance!
real-analysis convergence uniform-convergence
edited Jan 18 at 11:29
Bernard
121k740116
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
$endgroup$
I was wondering if provided that the partial sums $f_n$ converge to $f$ matching pointwise convergence, what this would imply:
$$
lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)
$$
If that would be the case, for all $x$, wouln't this imply uniform convergence? And if not, why.
Many thanks in advance!
real-analysis convergence uniform-convergence
real-analysis convergence uniform-convergence
edited Jan 18 at 11:29
Bernard
121k740116
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
edited Jan 18 at 11:29
Bernard
121k740116
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
edited Jan 18 at 11:29
Bernard
121k740116
edited Jan 18 at 11:29
Bernard
121k740116
edited Jan 18 at 11:29
Bernard
121k740116
121k740116
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
asked Jan 18 at 10:56
Carlos Toscano-OchoaCarlos Toscano-Ochoa
1,3971922
1,3971922
$begingroup$
So $f$ is a series?
$endgroup$
– Stockfish
Jan 18 at 10:58
1
$begingroup$
This is just saying $f$ is continuous. You won't necessarily have uniform convergence. Look at $f_n(x)=chi_{(0,1/n)}$ on $[0,1]$.
$endgroup$
– David Mitra
Jan 18 at 10:59
add a comment |
$begingroup$
So $f$ is a series?
$endgroup$
– Stockfish
Jan 18 at 10:58
1
$begingroup$
This is just saying $f$ is continuous. You won't necessarily have uniform convergence. Look at $f_n(x)=chi_{(0,1/n)}$ on $[0,1]$.
$endgroup$
– David Mitra
Jan 18 at 10:59
$begingroup$
So $f$ is a series?
$endgroup$
– Stockfish
Jan 18 at 10:58
$begingroup$
So $f$ is a series?
$endgroup$
– Stockfish
Jan 18 at 10:58
1
1
$begingroup$
This is just saying $f$ is continuous. You won't necessarily have uniform convergence. Look at $f_n(x)=chi_{(0,1/n)}$ on $[0,1]$.
$endgroup$
– David Mitra
Jan 18 at 10:59
$begingroup$
This is just saying $f$ is continuous. You won't necessarily have uniform convergence. Look at $f_n(x)=chi_{(0,1/n)}$ on $[0,1]$.
$endgroup$
– David Mitra
Jan 18 at 10:59
add a comment |
1 Answer
1
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Example: let $f_n(x)= frac{nx}{1+n^2x^2}$. Then $ (f_n)$ converges pointwise to the function $f(x)=0$ on $ mathbb R.$ Then we have $lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)$ for all $x$. But $(f_n)$ does not converge uniformly, since $f_n(1/n)=1/2$ for all $n$.
answered Jan 18 at 11:03
FredFred
46.9k1848
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1 Answer
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1 Answer
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$begingroup$
Example: let $f_n(x)= frac{nx}{1+n^2x^2}$. Then $ (f_n)$ converges pointwise to the function $f(x)=0$ on $ mathbb R.$ Then we have $lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)$ for all $x$. But $(f_n)$ does not converge uniformly, since $f_n(1/n)=1/2$ for all $n$.
answered Jan 18 at 11:03
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Example: let $f_n(x)= frac{nx}{1+n^2x^2}$. Then $ (f_n)$ converges pointwise to the function $f(x)=0$ on $ mathbb R.$ Then we have $lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)$ for all $x$. But $(f_n)$ does not converge uniformly, since $f_n(1/n)=1/2$ for all $n$.
answered Jan 18 at 11:03
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Example: let $f_n(x)= frac{nx}{1+n^2x^2}$. Then $ (f_n)$ converges pointwise to the function $f(x)=0$ on $ mathbb R.$ Then we have $lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)$ for all $x$. But $(f_n)$ does not converge uniformly, since $f_n(1/n)=1/2$ for all $n$.
answered Jan 18 at 11:03
FredFred
46.9k1848
$endgroup$
Example: let $f_n(x)= frac{nx}{1+n^2x^2}$. Then $ (f_n)$ converges pointwise to the function $f(x)=0$ on $ mathbb R.$ Then we have $lim_{varepsilon rightarrow 0} f(x+varepsilon) = f(x)$ for all $x$. But $(f_n)$ does not converge uniformly, since $f_n(1/n)=1/2$ for all $n$.
answered Jan 18 at 11:03
FredFred
46.9k1848
answered Jan 18 at 11:03
FredFred
46.9k1848
answered Jan 18 at 11:03
FredFred
46.9k1848
answered Jan 18 at 11:03
FredFred
46.9k1848
46.9k1848
add a comment |
add a comment |
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$begingroup$
So $f$ is a series?
$endgroup$
– Stockfish
Jan 18 at 10:58
1
$begingroup$
This is just saying $f$ is continuous. You won't necessarily have uniform convergence. Look at $f_n(x)=chi_{(0,1/n)}$ on $[0,1]$.
$endgroup$
– David Mitra
Jan 18 at 10:59