For each interval $[a,b]$ contained in $I$, sequence ${f_{n}:[a,b]rightarrowmathbb{R}}$ converges uniformly...
$begingroup$
$I$ is an open interval
Using the following fact to show this:
${{f_n}}$ converges pointwise on $I$ to the function $f$, and ${{f'_n}}$ converges uniformly on $I$ to the function $g$
Attempt:
$|{f'_n}(x)-g(x)|<epsilon/(2(b-a))$, and $|{f_n}(a)-f(a)|<epsilon/2$
Integrate the former equation from a to b to cancel out (b-a) so first equation $<epsilon/2$
Add the two and then $|{f_n}(x)-f(x)|<epsilon$
real-analysis sequences-and-series functions convergence
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
$endgroup$
add a comment |
$begingroup$
$I$ is an open interval
Using the following fact to show this:
${{f_n}}$ converges pointwise on $I$ to the function $f$, and ${{f'_n}}$ converges uniformly on $I$ to the function $g$
Attempt:
$|{f'_n}(x)-g(x)|<epsilon/(2(b-a))$, and $|{f_n}(a)-f(a)|<epsilon/2$
Integrate the former equation from a to b to cancel out (b-a) so first equation $<epsilon/2$
Add the two and then $|{f_n}(x)-f(x)|<epsilon$
real-analysis sequences-and-series functions convergence
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
$endgroup$
add a comment |
$begingroup$
$I$ is an open interval
Using the following fact to show this:
${{f_n}}$ converges pointwise on $I$ to the function $f$, and ${{f'_n}}$ converges uniformly on $I$ to the function $g$
Attempt:
$|{f'_n}(x)-g(x)|<epsilon/(2(b-a))$, and $|{f_n}(a)-f(a)|<epsilon/2$
Integrate the former equation from a to b to cancel out (b-a) so first equation $<epsilon/2$
Add the two and then $|{f_n}(x)-f(x)|<epsilon$
real-analysis sequences-and-series functions convergence
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
$endgroup$
$I$ is an open interval
Using the following fact to show this:
${{f_n}}$ converges pointwise on $I$ to the function $f$, and ${{f'_n}}$ converges uniformly on $I$ to the function $g$
Attempt:
$|{f'_n}(x)-g(x)|<epsilon/(2(b-a))$, and $|{f_n}(a)-f(a)|<epsilon/2$
Integrate the former equation from a to b to cancel out (b-a) so first equation $<epsilon/2$
Add the two and then $|{f_n}(x)-f(x)|<epsilon$
real-analysis sequences-and-series functions convergence
real-analysis sequences-and-series functions convergence
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
edited Dec 10 '18 at 7:14
Dillain Smith
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
asked Dec 10 '18 at 2:44
Dillain SmithDillain Smith
496
496
add a comment |
add a comment |
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