Discontinuity properties of $f_n$ carries over to the limit function $f$
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Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?
- No discontinuities
- At most ten discontinuities
- At least ten discontinuities
- Uncountably many discontinuities
- Countably many discontinuities
- No jump discontinuities
- No oscillating discontinuities
My try :
For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.
For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$
Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform
For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$
Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.
This link answers the sixth bullet
Is my arguments correct ? Can I have a hint for others ?
real-analysis sequence-of-function
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
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add a comment |
$begingroup$
Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?
- No discontinuities
- At most ten discontinuities
- At least ten discontinuities
- Uncountably many discontinuities
- Countably many discontinuities
- No jump discontinuities
- No oscillating discontinuities
My try :
For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.
For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$
Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform
For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$
Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.
This link answers the sixth bullet
Is my arguments correct ? Can I have a hint for others ?
real-analysis sequence-of-function
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
$endgroup$
1
$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55
$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35
add a comment |
$begingroup$
Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?
- No discontinuities
- At most ten discontinuities
- At least ten discontinuities
- Uncountably many discontinuities
- Countably many discontinuities
- No jump discontinuities
- No oscillating discontinuities
My try :
For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.
For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$
Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform
For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$
Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.
This link answers the sixth bullet
Is my arguments correct ? Can I have a hint for others ?
real-analysis sequence-of-function
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
$endgroup$
Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?
- No discontinuities
- At most ten discontinuities
- At least ten discontinuities
- Uncountably many discontinuities
- Countably many discontinuities
- No jump discontinuities
- No oscillating discontinuities
My try :
For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.
For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$
Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform
For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$
Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.
This link answers the sixth bullet
Is my arguments correct ? Can I have a hint for others ?
real-analysis sequence-of-function
real-analysis sequence-of-function
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
asked Dec 3 '18 at 18:20
Chinnapparaj RChinnapparaj R
5,3341828
5,3341828
1
$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55
$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35
add a comment |
1
$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55
$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35
1
1
$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55
$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55
$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35
$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35
add a comment |
1 Answer
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oldest
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$begingroup$
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
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add a comment |
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$begingroup$
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
$endgroup$
add a comment |
$begingroup$
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
$endgroup$
add a comment |
$begingroup$
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
$endgroup$
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
answered Dec 4 '18 at 8:52
IngixIngix
3,484146
3,484146
add a comment |
add a comment |
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$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55
$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35