Show that $f_n(x)=frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent
Show that $f_n(x)=frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent. For $f_n$ to be uniform convergent, $sup_{xinmathbb{R}}{|f_n(x)-f(x)|}ltepsilon$. I know that $f_n$ converges pointwise to $0$ if $ |x|leq1$, pointwise to $1/2$ if $x=pm 1$ and to $1$ otherwise. How can i formally prove, using above definition, that $f_n$ doest not converge uniformly? I know that the limit function is discontinuous and hence uniform convergence isn't possible but I would like a formal proof.
Thanks!
real-analysis sequences-and-series uniform-convergence
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
add a comment |
Show that $f_n(x)=frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent. For $f_n$ to be uniform convergent, $sup_{xinmathbb{R}}{|f_n(x)-f(x)|}ltepsilon$. I know that $f_n$ converges pointwise to $0$ if $ |x|leq1$, pointwise to $1/2$ if $x=pm 1$ and to $1$ otherwise. How can i formally prove, using above definition, that $f_n$ doest not converge uniformly? I know that the limit function is discontinuous and hence uniform convergence isn't possible but I would like a formal proof.
Thanks!
real-analysis sequences-and-series uniform-convergence
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
Byseq, what did you intend? I'm guessing $leq$, which isleq.
– J.G.
Dec 13 '18 at 10:42
Well, all the functions are continuous everywhere but they pointwise converge to a non-continuous function, so...
– DonAntonio
Dec 13 '18 at 10:45
corrected it, thanks
– Michael Maier
Dec 13 '18 at 10:45
2
When $vert x vert > 1$ [i.e., the "otherwise" case], isn't the limit $1$? Also uniform convergence is related to the domain, so you might need to add it. P.S. I think you want the proof for "non-uniform" convergence.
– xbh
Dec 13 '18 at 10:55
add a comment |
Show that $f_n(x)=frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent. For $f_n$ to be uniform convergent, $sup_{xinmathbb{R}}{|f_n(x)-f(x)|}ltepsilon$. I know that $f_n$ converges pointwise to $0$ if $ |x|leq1$, pointwise to $1/2$ if $x=pm 1$ and to $1$ otherwise. How can i formally prove, using above definition, that $f_n$ doest not converge uniformly? I know that the limit function is discontinuous and hence uniform convergence isn't possible but I would like a formal proof.
Thanks!
real-analysis sequences-and-series uniform-convergence
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
Show that $f_n(x)=frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent. For $f_n$ to be uniform convergent, $sup_{xinmathbb{R}}{|f_n(x)-f(x)|}ltepsilon$. I know that $f_n$ converges pointwise to $0$ if $ |x|leq1$, pointwise to $1/2$ if $x=pm 1$ and to $1$ otherwise. How can i formally prove, using above definition, that $f_n$ doest not converge uniformly? I know that the limit function is discontinuous and hence uniform convergence isn't possible but I would like a formal proof.
Thanks!
real-analysis sequences-and-series uniform-convergence
real-analysis sequences-and-series uniform-convergence
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
edited Dec 13 '18 at 11:45
Michael Maier
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
asked Dec 13 '18 at 10:38
Michael MaierMichael Maier
638
638
Byseq, what did you intend? I'm guessing $leq$, which isleq.
– J.G.
Dec 13 '18 at 10:42
Well, all the functions are continuous everywhere but they pointwise converge to a non-continuous function, so...
– DonAntonio
Dec 13 '18 at 10:45
corrected it, thanks
– Michael Maier
Dec 13 '18 at 10:45
2
When $vert x vert > 1$ [i.e., the "otherwise" case], isn't the limit $1$? Also uniform convergence is related to the domain, so you might need to add it. P.S. I think you want the proof for "non-uniform" convergence.
– xbh
Dec 13 '18 at 10:55
add a comment |
Byseq, what did you intend? I'm guessing $leq$, which isleq.
– J.G.
Dec 13 '18 at 10:42
Well, all the functions are continuous everywhere but they pointwise converge to a non-continuous function, so...
– DonAntonio
Dec 13 '18 at 10:45
corrected it, thanks
– Michael Maier
Dec 13 '18 at 10:45
2
When $vert x vert > 1$ [i.e., the "otherwise" case], isn't the limit $1$? Also uniform convergence is related to the domain, so you might need to add it. P.S. I think you want the proof for "non-uniform" convergence.
– xbh
Dec 13 '18 at 10:55
By
seq, what did you intend? I'm guessing $leq$, which is leq.– J.G.
Dec 13 '18 at 10:42
By
seq, what did you intend? I'm guessing $leq$, which is leq.– J.G.
Dec 13 '18 at 10:42
Well, all the functions are continuous everywhere but they pointwise converge to a non-continuous function, so...
– DonAntonio
Dec 13 '18 at 10:45
Well, all the functions are continuous everywhere but they pointwise converge to a non-continuous function, so...
– DonAntonio
Dec 13 '18 at 10:45
corrected it, thanks
– Michael Maier
Dec 13 '18 at 10:45
corrected it, thanks
– Michael Maier
Dec 13 '18 at 10:45
2
2
When $vert x vert > 1$ [i.e., the "otherwise" case], isn't the limit $1$? Also uniform convergence is related to the domain, so you might need to add it. P.S. I think you want the proof for "non-uniform" convergence.
– xbh
Dec 13 '18 at 10:55
When $vert x vert > 1$ [i.e., the "otherwise" case], isn't the limit $1$? Also uniform convergence is related to the domain, so you might need to add it. P.S. I think you want the proof for "non-uniform" convergence.
– xbh
Dec 13 '18 at 10:55
add a comment |
1 Answer
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Let $x_n=1+frac{1}{n}$. Then $x_n >1$, hence $|f_n(x_n)-f(x_n)|=f_n(x_n) to frac{e^2}{1+e^2}$. Now consider $epsilon >0$ with $epsilon < frac{e^2}{1+e^2}$.
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
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Let $x_n=1+frac{1}{n}$. Then $x_n >1$, hence $|f_n(x_n)-f(x_n)|=f_n(x_n) to frac{e^2}{1+e^2}$. Now consider $epsilon >0$ with $epsilon < frac{e^2}{1+e^2}$.
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
add a comment |
Let $x_n=1+frac{1}{n}$. Then $x_n >1$, hence $|f_n(x_n)-f(x_n)|=f_n(x_n) to frac{e^2}{1+e^2}$. Now consider $epsilon >0$ with $epsilon < frac{e^2}{1+e^2}$.
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
add a comment |
Let $x_n=1+frac{1}{n}$. Then $x_n >1$, hence $|f_n(x_n)-f(x_n)|=f_n(x_n) to frac{e^2}{1+e^2}$. Now consider $epsilon >0$ with $epsilon < frac{e^2}{1+e^2}$.
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
Let $x_n=1+frac{1}{n}$. Then $x_n >1$, hence $|f_n(x_n)-f(x_n)|=f_n(x_n) to frac{e^2}{1+e^2}$. Now consider $epsilon >0$ with $epsilon < frac{e^2}{1+e^2}$.
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
answered Dec 13 '18 at 11:39
FredFred
44.4k1845
answered Dec 13 '18 at 11:39
FredFred
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44.4k1845
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By
seq, what did you intend? I'm guessing $leq$, which isleq.– J.G.
Dec 13 '18 at 10:42
Well, all the functions are continuous everywhere but they pointwise converge to a non-continuous function, so...
– DonAntonio
Dec 13 '18 at 10:45
corrected it, thanks
– Michael Maier
Dec 13 '18 at 10:45
2
When $vert x vert > 1$ [i.e., the "otherwise" case], isn't the limit $1$? Also uniform convergence is related to the domain, so you might need to add it. P.S. I think you want the proof for "non-uniform" convergence.
– xbh
Dec 13 '18 at 10:55