For a branch-defined function, analyze the uniform convergence of it and of its series
Let $(f_n)$ be a sequence of functions on $mathbb{R}$ such that $$f_n(x) = left{ begin{array}{ll} 0, x leq 0, \ frac{x}{n^2}, x in (0,n^2), \ 1, x geq n^2. end{array}right. $$
a) Show that $(f_n)$ converges point-wise on $mathbb{R}$.
b) Determine whether $(f_n)$ converges uniformly on $mathbb{R}$.
c) Show that $displaystyle sum_{n=1}^infty f_n$ converges point-wise on $mathbb{R}$.
d) Show that $displaystyle sum_{n=1}^infty f_n$ is uniformly convergent on every interval $[0,a], a > 0$, but it is not uniformly convergent on $mathbb{R}$.
e) Show that $s:mathbb{R} to mathbb{R}$ given by $$ s(x) = sum_{n=1}^infty f_n(x)$$ is continuous.
For a), I think that we can easily see that $f_n$ tends to the function $f equiv 0$ for all $ x in mathbb{R}$ as $n to infty$.
For b), I am not sure how to determine whether $(f_n)$ is uniformly convergent. Do we just try and find out if $displaystyle sup_{x in mathbb{R}} {f_n(x) - 0 } to 0$ as $n to infty$? Because if that's how it should be solved, then the function does not converge uniformly, since the supremum is $1$.
I think that for c) we have that for all $x in mathbb{R}$, the series $displaystyle sum_{n =1}^infty f_n(x)$ contains a finite number of 1's and the other terms are $frac{x}{n^2},$ so I think it's safe to assume that the series converges point-wise?
I do not know how to start d) (maybe for the interval $[0,a]$, we can apply the Weierstrass M-test). My intuition says that there is something bad happening when $x to infty$. The same goes for e).
real-analysis sequences-and-series uniform-convergence
asked Nov 28 at 22:49
S.T.
174
add a comment |
Let $(f_n)$ be a sequence of functions on $mathbb{R}$ such that $$f_n(x) = left{ begin{array}{ll} 0, x leq 0, \ frac{x}{n^2}, x in (0,n^2), \ 1, x geq n^2. end{array}right. $$
a) Show that $(f_n)$ converges point-wise on $mathbb{R}$.
b) Determine whether $(f_n)$ converges uniformly on $mathbb{R}$.
c) Show that $displaystyle sum_{n=1}^infty f_n$ converges point-wise on $mathbb{R}$.
d) Show that $displaystyle sum_{n=1}^infty f_n$ is uniformly convergent on every interval $[0,a], a > 0$, but it is not uniformly convergent on $mathbb{R}$.
e) Show that $s:mathbb{R} to mathbb{R}$ given by $$ s(x) = sum_{n=1}^infty f_n(x)$$ is continuous.
For a), I think that we can easily see that $f_n$ tends to the function $f equiv 0$ for all $ x in mathbb{R}$ as $n to infty$.
For b), I am not sure how to determine whether $(f_n)$ is uniformly convergent. Do we just try and find out if $displaystyle sup_{x in mathbb{R}} {f_n(x) - 0 } to 0$ as $n to infty$? Because if that's how it should be solved, then the function does not converge uniformly, since the supremum is $1$.
I think that for c) we have that for all $x in mathbb{R}$, the series $displaystyle sum_{n =1}^infty f_n(x)$ contains a finite number of 1's and the other terms are $frac{x}{n^2},$ so I think it's safe to assume that the series converges point-wise?
I do not know how to start d) (maybe for the interval $[0,a]$, we can apply the Weierstrass M-test). My intuition says that there is something bad happening when $x to infty$. The same goes for e).
real-analysis sequences-and-series uniform-convergence
asked Nov 28 at 22:49
S.T.
174
Hint for $b.)$ what is $f_n(n^2)-f(n^2)$?
– Severin Schraven
Nov 28 at 22:52
That's exactly what I thought, I also edited the question right about now.
– S.T.
Nov 28 at 22:53
Uniform convergence doesn't mean for $sum_{n=1}^infty f_n(x)$. Possibly you mean $sum_{k=1}^n f_k(x)$ right?
– Mostafa Ayaz
Nov 29 at 8:43
add a comment |
Let $(f_n)$ be a sequence of functions on $mathbb{R}$ such that $$f_n(x) = left{ begin{array}{ll} 0, x leq 0, \ frac{x}{n^2}, x in (0,n^2), \ 1, x geq n^2. end{array}right. $$
a) Show that $(f_n)$ converges point-wise on $mathbb{R}$.
b) Determine whether $(f_n)$ converges uniformly on $mathbb{R}$.
c) Show that $displaystyle sum_{n=1}^infty f_n$ converges point-wise on $mathbb{R}$.
d) Show that $displaystyle sum_{n=1}^infty f_n$ is uniformly convergent on every interval $[0,a], a > 0$, but it is not uniformly convergent on $mathbb{R}$.
e) Show that $s:mathbb{R} to mathbb{R}$ given by $$ s(x) = sum_{n=1}^infty f_n(x)$$ is continuous.
For a), I think that we can easily see that $f_n$ tends to the function $f equiv 0$ for all $ x in mathbb{R}$ as $n to infty$.
For b), I am not sure how to determine whether $(f_n)$ is uniformly convergent. Do we just try and find out if $displaystyle sup_{x in mathbb{R}} {f_n(x) - 0 } to 0$ as $n to infty$? Because if that's how it should be solved, then the function does not converge uniformly, since the supremum is $1$.
I think that for c) we have that for all $x in mathbb{R}$, the series $displaystyle sum_{n =1}^infty f_n(x)$ contains a finite number of 1's and the other terms are $frac{x}{n^2},$ so I think it's safe to assume that the series converges point-wise?
I do not know how to start d) (maybe for the interval $[0,a]$, we can apply the Weierstrass M-test). My intuition says that there is something bad happening when $x to infty$. The same goes for e).
real-analysis sequences-and-series uniform-convergence
asked Nov 28 at 22:49
S.T.
174
Let $(f_n)$ be a sequence of functions on $mathbb{R}$ such that $$f_n(x) = left{ begin{array}{ll} 0, x leq 0, \ frac{x}{n^2}, x in (0,n^2), \ 1, x geq n^2. end{array}right. $$
a) Show that $(f_n)$ converges point-wise on $mathbb{R}$.
b) Determine whether $(f_n)$ converges uniformly on $mathbb{R}$.
c) Show that $displaystyle sum_{n=1}^infty f_n$ converges point-wise on $mathbb{R}$.
d) Show that $displaystyle sum_{n=1}^infty f_n$ is uniformly convergent on every interval $[0,a], a > 0$, but it is not uniformly convergent on $mathbb{R}$.
e) Show that $s:mathbb{R} to mathbb{R}$ given by $$ s(x) = sum_{n=1}^infty f_n(x)$$ is continuous.
For a), I think that we can easily see that $f_n$ tends to the function $f equiv 0$ for all $ x in mathbb{R}$ as $n to infty$.
For b), I am not sure how to determine whether $(f_n)$ is uniformly convergent. Do we just try and find out if $displaystyle sup_{x in mathbb{R}} {f_n(x) - 0 } to 0$ as $n to infty$? Because if that's how it should be solved, then the function does not converge uniformly, since the supremum is $1$.
I think that for c) we have that for all $x in mathbb{R}$, the series $displaystyle sum_{n =1}^infty f_n(x)$ contains a finite number of 1's and the other terms are $frac{x}{n^2},$ so I think it's safe to assume that the series converges point-wise?
I do not know how to start d) (maybe for the interval $[0,a]$, we can apply the Weierstrass M-test). My intuition says that there is something bad happening when $x to infty$. The same goes for e).
real-analysis sequences-and-series uniform-convergence
real-analysis sequences-and-series uniform-convergence
asked Nov 28 at 22:49
S.T.
174
asked Nov 28 at 22:49
S.T.
174
edited Nov 28 at 22:53
asked Nov 28 at 22:49
S.T.
174
asked Nov 28 at 22:49
S.T.
174
asked Nov 28 at 22:49
S.T.
174
174
Hint for $b.)$ what is $f_n(n^2)-f(n^2)$?
– Severin Schraven
Nov 28 at 22:52
That's exactly what I thought, I also edited the question right about now.
– S.T.
Nov 28 at 22:53
Uniform convergence doesn't mean for $sum_{n=1}^infty f_n(x)$. Possibly you mean $sum_{k=1}^n f_k(x)$ right?
– Mostafa Ayaz
Nov 29 at 8:43
add a comment |
Hint for $b.)$ what is $f_n(n^2)-f(n^2)$?
– Severin Schraven
Nov 28 at 22:52
That's exactly what I thought, I also edited the question right about now.
– S.T.
Nov 28 at 22:53
Uniform convergence doesn't mean for $sum_{n=1}^infty f_n(x)$. Possibly you mean $sum_{k=1}^n f_k(x)$ right?
– Mostafa Ayaz
Nov 29 at 8:43
Hint for $b.)$ what is $f_n(n^2)-f(n^2)$?
– Severin Schraven
Nov 28 at 22:52
Hint for $b.)$ what is $f_n(n^2)-f(n^2)$?
– Severin Schraven
Nov 28 at 22:52
That's exactly what I thought, I also edited the question right about now.
– S.T.
Nov 28 at 22:53
That's exactly what I thought, I also edited the question right about now.
– S.T.
Nov 28 at 22:53
Uniform convergence doesn't mean for $sum_{n=1}^infty f_n(x)$. Possibly you mean $sum_{k=1}^n f_k(x)$ right?
– Mostafa Ayaz
Nov 29 at 8:43
Uniform convergence doesn't mean for $sum_{n=1}^infty f_n(x)$. Possibly you mean $sum_{k=1}^n f_k(x)$ right?
– Mostafa Ayaz
Nov 29 at 8:43
add a comment |
2 Answers
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(b) $f_n$ is not uniformly convergent. We show that this is not the case even when $xin (0,n^2)$. Recall the definition of uniform convergence:$$forall epsilon>0quad,quad exists Nquad,quadforall n>Nquad,quad |{xover n^2}|<epsilon$$therefore $$n>{sqrt{xover epsilon}}$$this means that we must choose $N=lfloor {sqrt{xover epsilon}}rfloor +1$ which is surely dependent to $x$ not only to $epsilon$, therefore $f_n(x)$ is not uniformly convergent.
(c) the statement is false since $s(x)$ is no function of $n$, but for $s_n(x)$ with the following definition$$s_n(x)=sum_{k=1}^{n} f_k(x)$$we have $$s_n(x)to s(x)$$where $$s(x)=sum_{k=1}^{infty} f_k(x)=begin{cases}0&,quad x<0\sum_{n=1\ n>sqrt x}^{infty}{xover n^2}+sum_{n=1\nle sqrt x}1&,quad xge 0end{cases}$$here is a sketch of $s(x)$
As you see, the function is continuous (which can be proved easily only in points $x$ with $x=k^2$ for some $kin Bbb Z$) and piecewise linear whose slope on each piece (from $k^2$ to $(k+1)^2$ increases unboundedly) therefore the function is uniformly continuous on each $(0,a)$ for $ain Bbb R^+$but not on $Bbb R$. The proofs are easy and straight forward. Finally we have
d,e) $s(x)$ is continuous though not uniformly and $s_n(x)$ tends to $s(x)$ pointwise
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
add a comment |
I think e) is false. In fact, $s(1+frac 1 k) to sum_{n=2}^{infty} frac 1 {n^{2}}+1$ whereas $s(1) = sum_{n=2}^{infty} frac 1 {n^{2}}$.
Your argument for a) is correct. For c) use the fact that $f_n(x) leq frac x {n^{2}}$ for all $x geq 0$. This also proves uniform convergence on $[0,a]$ by M-test. Second part of d) follows from b).
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
add a comment |
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2 Answers
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2 Answers
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(b) $f_n$ is not uniformly convergent. We show that this is not the case even when $xin (0,n^2)$. Recall the definition of uniform convergence:$$forall epsilon>0quad,quad exists Nquad,quadforall n>Nquad,quad |{xover n^2}|<epsilon$$therefore $$n>{sqrt{xover epsilon}}$$this means that we must choose $N=lfloor {sqrt{xover epsilon}}rfloor +1$ which is surely dependent to $x$ not only to $epsilon$, therefore $f_n(x)$ is not uniformly convergent.
(c) the statement is false since $s(x)$ is no function of $n$, but for $s_n(x)$ with the following definition$$s_n(x)=sum_{k=1}^{n} f_k(x)$$we have $$s_n(x)to s(x)$$where $$s(x)=sum_{k=1}^{infty} f_k(x)=begin{cases}0&,quad x<0\sum_{n=1\ n>sqrt x}^{infty}{xover n^2}+sum_{n=1\nle sqrt x}1&,quad xge 0end{cases}$$here is a sketch of $s(x)$
As you see, the function is continuous (which can be proved easily only in points $x$ with $x=k^2$ for some $kin Bbb Z$) and piecewise linear whose slope on each piece (from $k^2$ to $(k+1)^2$ increases unboundedly) therefore the function is uniformly continuous on each $(0,a)$ for $ain Bbb R^+$but not on $Bbb R$. The proofs are easy and straight forward. Finally we have
d,e) $s(x)$ is continuous though not uniformly and $s_n(x)$ tends to $s(x)$ pointwise
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
add a comment |
(b) $f_n$ is not uniformly convergent. We show that this is not the case even when $xin (0,n^2)$. Recall the definition of uniform convergence:$$forall epsilon>0quad,quad exists Nquad,quadforall n>Nquad,quad |{xover n^2}|<epsilon$$therefore $$n>{sqrt{xover epsilon}}$$this means that we must choose $N=lfloor {sqrt{xover epsilon}}rfloor +1$ which is surely dependent to $x$ not only to $epsilon$, therefore $f_n(x)$ is not uniformly convergent.
(c) the statement is false since $s(x)$ is no function of $n$, but for $s_n(x)$ with the following definition$$s_n(x)=sum_{k=1}^{n} f_k(x)$$we have $$s_n(x)to s(x)$$where $$s(x)=sum_{k=1}^{infty} f_k(x)=begin{cases}0&,quad x<0\sum_{n=1\ n>sqrt x}^{infty}{xover n^2}+sum_{n=1\nle sqrt x}1&,quad xge 0end{cases}$$here is a sketch of $s(x)$
As you see, the function is continuous (which can be proved easily only in points $x$ with $x=k^2$ for some $kin Bbb Z$) and piecewise linear whose slope on each piece (from $k^2$ to $(k+1)^2$ increases unboundedly) therefore the function is uniformly continuous on each $(0,a)$ for $ain Bbb R^+$but not on $Bbb R$. The proofs are easy and straight forward. Finally we have
d,e) $s(x)$ is continuous though not uniformly and $s_n(x)$ tends to $s(x)$ pointwise
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
add a comment |
(b) $f_n$ is not uniformly convergent. We show that this is not the case even when $xin (0,n^2)$. Recall the definition of uniform convergence:$$forall epsilon>0quad,quad exists Nquad,quadforall n>Nquad,quad |{xover n^2}|<epsilon$$therefore $$n>{sqrt{xover epsilon}}$$this means that we must choose $N=lfloor {sqrt{xover epsilon}}rfloor +1$ which is surely dependent to $x$ not only to $epsilon$, therefore $f_n(x)$ is not uniformly convergent.
(c) the statement is false since $s(x)$ is no function of $n$, but for $s_n(x)$ with the following definition$$s_n(x)=sum_{k=1}^{n} f_k(x)$$we have $$s_n(x)to s(x)$$where $$s(x)=sum_{k=1}^{infty} f_k(x)=begin{cases}0&,quad x<0\sum_{n=1\ n>sqrt x}^{infty}{xover n^2}+sum_{n=1\nle sqrt x}1&,quad xge 0end{cases}$$here is a sketch of $s(x)$
As you see, the function is continuous (which can be proved easily only in points $x$ with $x=k^2$ for some $kin Bbb Z$) and piecewise linear whose slope on each piece (from $k^2$ to $(k+1)^2$ increases unboundedly) therefore the function is uniformly continuous on each $(0,a)$ for $ain Bbb R^+$but not on $Bbb R$. The proofs are easy and straight forward. Finally we have
d,e) $s(x)$ is continuous though not uniformly and $s_n(x)$ tends to $s(x)$ pointwise
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
(b) $f_n$ is not uniformly convergent. We show that this is not the case even when $xin (0,n^2)$. Recall the definition of uniform convergence:$$forall epsilon>0quad,quad exists Nquad,quadforall n>Nquad,quad |{xover n^2}|<epsilon$$therefore $$n>{sqrt{xover epsilon}}$$this means that we must choose $N=lfloor {sqrt{xover epsilon}}rfloor +1$ which is surely dependent to $x$ not only to $epsilon$, therefore $f_n(x)$ is not uniformly convergent.
(c) the statement is false since $s(x)$ is no function of $n$, but for $s_n(x)$ with the following definition$$s_n(x)=sum_{k=1}^{n} f_k(x)$$we have $$s_n(x)to s(x)$$where $$s(x)=sum_{k=1}^{infty} f_k(x)=begin{cases}0&,quad x<0\sum_{n=1\ n>sqrt x}^{infty}{xover n^2}+sum_{n=1\nle sqrt x}1&,quad xge 0end{cases}$$here is a sketch of $s(x)$
As you see, the function is continuous (which can be proved easily only in points $x$ with $x=k^2$ for some $kin Bbb Z$) and piecewise linear whose slope on each piece (from $k^2$ to $(k+1)^2$ increases unboundedly) therefore the function is uniformly continuous on each $(0,a)$ for $ain Bbb R^+$but not on $Bbb R$. The proofs are easy and straight forward. Finally we have
d,e) $s(x)$ is continuous though not uniformly and $s_n(x)$ tends to $s(x)$ pointwise
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
answered Nov 29 at 9:15
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
I think e) is false. In fact, $s(1+frac 1 k) to sum_{n=2}^{infty} frac 1 {n^{2}}+1$ whereas $s(1) = sum_{n=2}^{infty} frac 1 {n^{2}}$.
Your argument for a) is correct. For c) use the fact that $f_n(x) leq frac x {n^{2}}$ for all $x geq 0$. This also proves uniform convergence on $[0,a]$ by M-test. Second part of d) follows from b).
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
add a comment |
I think e) is false. In fact, $s(1+frac 1 k) to sum_{n=2}^{infty} frac 1 {n^{2}}+1$ whereas $s(1) = sum_{n=2}^{infty} frac 1 {n^{2}}$.
Your argument for a) is correct. For c) use the fact that $f_n(x) leq frac x {n^{2}}$ for all $x geq 0$. This also proves uniform convergence on $[0,a]$ by M-test. Second part of d) follows from b).
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
add a comment |
I think e) is false. In fact, $s(1+frac 1 k) to sum_{n=2}^{infty} frac 1 {n^{2}}+1$ whereas $s(1) = sum_{n=2}^{infty} frac 1 {n^{2}}$.
Your argument for a) is correct. For c) use the fact that $f_n(x) leq frac x {n^{2}}$ for all $x geq 0$. This also proves uniform convergence on $[0,a]$ by M-test. Second part of d) follows from b).
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
I think e) is false. In fact, $s(1+frac 1 k) to sum_{n=2}^{infty} frac 1 {n^{2}}+1$ whereas $s(1) = sum_{n=2}^{infty} frac 1 {n^{2}}$.
Your argument for a) is correct. For c) use the fact that $f_n(x) leq frac x {n^{2}}$ for all $x geq 0$. This also proves uniform convergence on $[0,a]$ by M-test. Second part of d) follows from b).
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
answered Nov 28 at 23:38
Kavi Rama Murthy
48.6k31854
48.6k31854
add a comment |
add a comment |
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Hint for $b.)$ what is $f_n(n^2)-f(n^2)$?
– Severin Schraven
Nov 28 at 22:52
That's exactly what I thought, I also edited the question right about now.
– S.T.
Nov 28 at 22:53
Uniform convergence doesn't mean for $sum_{n=1}^infty f_n(x)$. Possibly you mean $sum_{k=1}^n f_k(x)$ right?
– Mostafa Ayaz
Nov 29 at 8:43