uniformly convergence
suppose $f_n(x)$ is a sequence of complex functions which converges uniformly to $f(x)$,$g_n(x)$ is a sequence of complex functions such that $|f_n(x)-g_n(x)|to 0$.Can we conclude that $g_n(x)$ is uniformly convergent to $f(x)$?
calculus real-analysis complex-analysis analysis uniform-convergence
asked Nov 30 at 0:33
mathrookie
800512
add a comment |
suppose $f_n(x)$ is a sequence of complex functions which converges uniformly to $f(x)$,$g_n(x)$ is a sequence of complex functions such that $|f_n(x)-g_n(x)|to 0$.Can we conclude that $g_n(x)$ is uniformly convergent to $f(x)$?
calculus real-analysis complex-analysis analysis uniform-convergence
asked Nov 30 at 0:33
mathrookie
800512
1
What is $||cdot||$?
– Guacho Perez
Nov 30 at 0:42
add a comment |
suppose $f_n(x)$ is a sequence of complex functions which converges uniformly to $f(x)$,$g_n(x)$ is a sequence of complex functions such that $|f_n(x)-g_n(x)|to 0$.Can we conclude that $g_n(x)$ is uniformly convergent to $f(x)$?
calculus real-analysis complex-analysis analysis uniform-convergence
asked Nov 30 at 0:33
mathrookie
800512
suppose $f_n(x)$ is a sequence of complex functions which converges uniformly to $f(x)$,$g_n(x)$ is a sequence of complex functions such that $|f_n(x)-g_n(x)|to 0$.Can we conclude that $g_n(x)$ is uniformly convergent to $f(x)$?
calculus real-analysis complex-analysis analysis uniform-convergence
calculus real-analysis complex-analysis analysis uniform-convergence
asked Nov 30 at 0:33
mathrookie
800512
asked Nov 30 at 0:33
mathrookie
800512
edited Nov 30 at 2:10
asked Nov 30 at 0:33
mathrookie
800512
asked Nov 30 at 0:33
mathrookie
800512
asked Nov 30 at 0:33
mathrookie
800512
800512
1
What is $||cdot||$?
– Guacho Perez
Nov 30 at 0:42
add a comment |
1
What is $||cdot||$?
– Guacho Perez
Nov 30 at 0:42
1
1
What is $||cdot||$?
– Guacho Perez
Nov 30 at 0:42
What is $||cdot||$?
– Guacho Perez
Nov 30 at 0:42
add a comment |
1 Answer
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Unless $lvert{f_n(x) - g_n(x)}rvert rightarrow 0$ is uniform, no. For instance, take $f_n(x) equiv 0$ and $f equiv 0$. Then $f_n rightarrow f$ uniformly on $(0, 1) subset mathbb{R} subset mathbb{C}$. Take $g_n(x) = x^n$. Then $lvert{g_n(x)rvert} rightarrow 0$ for all $x in (0, 1)$ pointwise but not uniformly.
If $lvert{f_n(x) - g_n(x)rvert} rightarrow 0$ uniformly, then your statement is true as $$lvert{g_n(x) - f(x)rvert} leq lvert{g_n(x) - f_n(x)rvert} + lvert{f_n(x) - f(x)}rvert$$ and you can bound both terms uniformly by assumption.
answered Nov 30 at 1:22
Riley
1245
add a comment |
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Unless $lvert{f_n(x) - g_n(x)}rvert rightarrow 0$ is uniform, no. For instance, take $f_n(x) equiv 0$ and $f equiv 0$. Then $f_n rightarrow f$ uniformly on $(0, 1) subset mathbb{R} subset mathbb{C}$. Take $g_n(x) = x^n$. Then $lvert{g_n(x)rvert} rightarrow 0$ for all $x in (0, 1)$ pointwise but not uniformly.
If $lvert{f_n(x) - g_n(x)rvert} rightarrow 0$ uniformly, then your statement is true as $$lvert{g_n(x) - f(x)rvert} leq lvert{g_n(x) - f_n(x)rvert} + lvert{f_n(x) - f(x)}rvert$$ and you can bound both terms uniformly by assumption.
answered Nov 30 at 1:22
Riley
1245
add a comment |
Unless $lvert{f_n(x) - g_n(x)}rvert rightarrow 0$ is uniform, no. For instance, take $f_n(x) equiv 0$ and $f equiv 0$. Then $f_n rightarrow f$ uniformly on $(0, 1) subset mathbb{R} subset mathbb{C}$. Take $g_n(x) = x^n$. Then $lvert{g_n(x)rvert} rightarrow 0$ for all $x in (0, 1)$ pointwise but not uniformly.
If $lvert{f_n(x) - g_n(x)rvert} rightarrow 0$ uniformly, then your statement is true as $$lvert{g_n(x) - f(x)rvert} leq lvert{g_n(x) - f_n(x)rvert} + lvert{f_n(x) - f(x)}rvert$$ and you can bound both terms uniformly by assumption.
answered Nov 30 at 1:22
Riley
1245
add a comment |
Unless $lvert{f_n(x) - g_n(x)}rvert rightarrow 0$ is uniform, no. For instance, take $f_n(x) equiv 0$ and $f equiv 0$. Then $f_n rightarrow f$ uniformly on $(0, 1) subset mathbb{R} subset mathbb{C}$. Take $g_n(x) = x^n$. Then $lvert{g_n(x)rvert} rightarrow 0$ for all $x in (0, 1)$ pointwise but not uniformly.
If $lvert{f_n(x) - g_n(x)rvert} rightarrow 0$ uniformly, then your statement is true as $$lvert{g_n(x) - f(x)rvert} leq lvert{g_n(x) - f_n(x)rvert} + lvert{f_n(x) - f(x)}rvert$$ and you can bound both terms uniformly by assumption.
answered Nov 30 at 1:22
Riley
1245
Unless $lvert{f_n(x) - g_n(x)}rvert rightarrow 0$ is uniform, no. For instance, take $f_n(x) equiv 0$ and $f equiv 0$. Then $f_n rightarrow f$ uniformly on $(0, 1) subset mathbb{R} subset mathbb{C}$. Take $g_n(x) = x^n$. Then $lvert{g_n(x)rvert} rightarrow 0$ for all $x in (0, 1)$ pointwise but not uniformly.
If $lvert{f_n(x) - g_n(x)rvert} rightarrow 0$ uniformly, then your statement is true as $$lvert{g_n(x) - f(x)rvert} leq lvert{g_n(x) - f_n(x)rvert} + lvert{f_n(x) - f(x)}rvert$$ and you can bound both terms uniformly by assumption.
answered Nov 30 at 1:22
Riley
1245
answered Nov 30 at 1:22
Riley
1245
answered Nov 30 at 1:22
Riley
1245
answered Nov 30 at 1:22
Riley
1245
1245
add a comment |
add a comment |
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1
What is $||cdot||$?
– Guacho Perez
Nov 30 at 0:42