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0

$begingroup$

Let $f_n ,,fcolon [0,1]rightarrow [0,infty)$ be given.

I wanted to show that if $f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{1/2}_{n}{underset {n }{rightrightarrows }}f^{1/2}$

Using $|sqrt{a}-sqrt{b}|le sqrt{|a-b|}$ we see that
$$lim_{nto infty}sup_{xin[0,1]}left | sqrt{f_n(x)}-sqrt{f(x)}right | le lim_{nto infty}sup_{xin[0,1]}sqrt{left | f_n(x)-f(x)right |}le
lim_{nto infty}sup_{xin[0,1]}left | f_n(x)-f(x)right |=0$$

and the claim follows.

Is this correct?

I however can not find an example where

$f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$

This cannot be true I think, can it?

real-analysis uniform-convergence

share|cite|improve this question

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

$endgroup$

add a comment | 

0

$begingroup$

Let $f_n ,,fcolon [0,1]rightarrow [0,infty)$ be given.

I wanted to show that if $f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{1/2}_{n}{underset {n }{rightrightarrows }}f^{1/2}$

Using $|sqrt{a}-sqrt{b}|le sqrt{|a-b|}$ we see that
$$lim_{nto infty}sup_{xin[0,1]}left | sqrt{f_n(x)}-sqrt{f(x)}right | le lim_{nto infty}sup_{xin[0,1]}sqrt{left | f_n(x)-f(x)right |}le
lim_{nto infty}sup_{xin[0,1]}left | f_n(x)-f(x)right |=0$$

and the claim follows.

Is this correct?

I however can not find an example where

$f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$

This cannot be true I think, can it?

real-analysis uniform-convergence

share|cite|improve this question

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

$endgroup$

add a comment | 

$begingroup$

Let $f_n ,,fcolon [0,1]rightarrow [0,infty)$ be given.

I wanted to show that if $f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{1/2}_{n}{underset {n }{rightrightarrows }}f^{1/2}$

Using $|sqrt{a}-sqrt{b}|le sqrt{|a-b|}$ we see that
$$lim_{nto infty}sup_{xin[0,1]}left | sqrt{f_n(x)}-sqrt{f(x)}right | le lim_{nto infty}sup_{xin[0,1]}sqrt{left | f_n(x)-f(x)right |}le
lim_{nto infty}sup_{xin[0,1]}left | f_n(x)-f(x)right |=0$$

and the claim follows.

Is this correct?

I however can not find an example where

$f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$

This cannot be true I think, can it?

real-analysis uniform-convergence

share|cite|improve this question

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

$endgroup$

share|cite|improve this question

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

share|cite|improve this question

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

asked Mar 9 at 16:47

EpsilonDeltaEpsilonDelta

6921615

1 Answer
1

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2

$begingroup$

In the first part, the claim $f_{n}{underset {n }{rightrightarrows }}f implies f_{n}^{1/2}{underset {n }{rightrightarrows }}f^{1/2} $ for $f_n,f :[0,1] to [0,infty)$ is true. Your proof starts out correctly, but your last inequality is not valid. You can complete the argument by stating that for all $epsilon > 0$ there exists $N$ independent of $x$ such that when $n > N$ we have for all $x in [0,1]$,

$$|f_n(x) - f(x)| < epsilon^2, $$

and, thus,

$$left|sqrt{f_n(x)} - sqrt{f(x)}right| leqslant sqrt{|f_n(x) - f(x)|} < epsilon.$$

Second Part

Clearly, there are examples where both $f_{n}{underset {n }{rightrightarrows }}f$ and $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ hold, either by virtue of your first proposition or the trivial example $f_n(x) = 1/n, , f(x) = 0$.

However, the implication$f_{n}{underset {n }{rightrightarrows }}fimplies f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ is not true.

For a counterexample, take $f_n,f:(0,1] to mathbb{R}$ where

$$f_n(x) = log frac{nx}{n+1}, quad f(x) = log x$$

We have $f_{n}{underset {n }{rightrightarrows }}f$, since

$$|f_n(x) - f(x)| = log frac{n+1}{n} to 0$$

However,
$$|f_n^2(x) - f^2(x)| = left|log frac{nx}{n+1 } - log xright|left|log frac{nx}{n+1 } + log xright| = logfrac{n+1}{n} left|log frac{nx^2}{n+1 }right|,$$

and by choosing $x_n = 1/(n e^n)$ we have

$$|f_n^2(x_n) - f^2(x_n)| = logfrac{n+1}{n} log (n^2 + n) + 2n logfrac{n+1}{n} $$

Since the RHS does not converge to $0$ as $n to infty$, it follows that $f_{n}^2{underset {n }{rightrightarrows }}f^2$ does not hold on $(0,1].$

This example is easily modified for the domain $[0,1]$ and range $[0,infty)$ as $f_n(x) = -log frac{nx}{n+1} mathbf{1}_{(0,1]}.$

share|cite|improve this answer

edited Mar 9 at 23:33

answered Mar 9 at 22:02

RRLRRL

52.5k42573

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi RRL. This doesn't seem to address the question. If $f_n^2$ is UC, is $f_n$ UC?
  $endgroup$
  – Mark Viola
  Mar 9 at 22:33
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkViola no, in fact this is on topic, read the title and the second last sentence.
  $endgroup$
  – enedil
  Mar 9 at 22:41
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The wording of the second part is not the best. "I however can not find an example where $f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$." If this is not an implication -- i.e. read "and" in place of "then" --then I showed both can hold. "This cannot be true I think, can it?" If by this OP is doubting the implication, then I showed it is false with a counterexample.
  $endgroup$
  – RRL
  Mar 9 at 23:14
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

In the first part, the claim $f_{n}{underset {n }{rightrightarrows }}f implies f_{n}^{1/2}{underset {n }{rightrightarrows }}f^{1/2} $ for $f_n,f :[0,1] to [0,infty)$ is true. Your proof starts out correctly, but your last inequality is not valid. You can complete the argument by stating that for all $epsilon > 0$ there exists $N$ independent of $x$ such that when $n > N$ we have for all $x in [0,1]$,

$$|f_n(x) - f(x)| < epsilon^2, $$

and, thus,

$$left|sqrt{f_n(x)} - sqrt{f(x)}right| leqslant sqrt{|f_n(x) - f(x)|} < epsilon.$$

Second Part

Clearly, there are examples where both $f_{n}{underset {n }{rightrightarrows }}f$ and $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ hold, either by virtue of your first proposition or the trivial example $f_n(x) = 1/n, , f(x) = 0$.

However, the implication$f_{n}{underset {n }{rightrightarrows }}fimplies f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ is not true.

For a counterexample, take $f_n,f:(0,1] to mathbb{R}$ where

$$f_n(x) = log frac{nx}{n+1}, quad f(x) = log x$$

We have $f_{n}{underset {n }{rightrightarrows }}f$, since

$$|f_n(x) - f(x)| = log frac{n+1}{n} to 0$$

However,
$$|f_n^2(x) - f^2(x)| = left|log frac{nx}{n+1 } - log xright|left|log frac{nx}{n+1 } + log xright| = logfrac{n+1}{n} left|log frac{nx^2}{n+1 }right|,$$

and by choosing $x_n = 1/(n e^n)$ we have

$$|f_n^2(x_n) - f^2(x_n)| = logfrac{n+1}{n} log (n^2 + n) + 2n logfrac{n+1}{n} $$

Since the RHS does not converge to $0$ as $n to infty$, it follows that $f_{n}^2{underset {n }{rightrightarrows }}f^2$ does not hold on $(0,1].$

This example is easily modified for the domain $[0,1]$ and range $[0,infty)$ as $f_n(x) = -log frac{nx}{n+1} mathbf{1}_{(0,1]}.$

share|cite|improve this answer

edited Mar 9 at 23:33

answered Mar 9 at 22:02

RRLRRL

52.5k42573

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi RRL. This doesn't seem to address the question. If $f_n^2$ is UC, is $f_n$ UC?
  $endgroup$
  – Mark Viola
  Mar 9 at 22:33
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkViola no, in fact this is on topic, read the title and the second last sentence.
  $endgroup$
  – enedil
  Mar 9 at 22:41
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The wording of the second part is not the best. "I however can not find an example where $f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$." If this is not an implication -- i.e. read "and" in place of "then" --then I showed both can hold. "This cannot be true I think, can it?" If by this OP is doubting the implication, then I showed it is false with a counterexample.
  $endgroup$
  – RRL
  Mar 9 at 23:14
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

In the first part, the claim $f_{n}{underset {n }{rightrightarrows }}f implies f_{n}^{1/2}{underset {n }{rightrightarrows }}f^{1/2} $ for $f_n,f :[0,1] to [0,infty)$ is true. Your proof starts out correctly, but your last inequality is not valid. You can complete the argument by stating that for all $epsilon > 0$ there exists $N$ independent of $x$ such that when $n > N$ we have for all $x in [0,1]$,

$$|f_n(x) - f(x)| < epsilon^2, $$

and, thus,

$$left|sqrt{f_n(x)} - sqrt{f(x)}right| leqslant sqrt{|f_n(x) - f(x)|} < epsilon.$$

Second Part

Clearly, there are examples where both $f_{n}{underset {n }{rightrightarrows }}f$ and $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ hold, either by virtue of your first proposition or the trivial example $f_n(x) = 1/n, , f(x) = 0$.

However, the implication$f_{n}{underset {n }{rightrightarrows }}fimplies f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ is not true.

For a counterexample, take $f_n,f:(0,1] to mathbb{R}$ where

$$f_n(x) = log frac{nx}{n+1}, quad f(x) = log x$$

We have $f_{n}{underset {n }{rightrightarrows }}f$, since

$$|f_n(x) - f(x)| = log frac{n+1}{n} to 0$$

However,
$$|f_n^2(x) - f^2(x)| = left|log frac{nx}{n+1 } - log xright|left|log frac{nx}{n+1 } + log xright| = logfrac{n+1}{n} left|log frac{nx^2}{n+1 }right|,$$

and by choosing $x_n = 1/(n e^n)$ we have

$$|f_n^2(x_n) - f^2(x_n)| = logfrac{n+1}{n} log (n^2 + n) + 2n logfrac{n+1}{n} $$

Since the RHS does not converge to $0$ as $n to infty$, it follows that $f_{n}^2{underset {n }{rightrightarrows }}f^2$ does not hold on $(0,1].$

This example is easily modified for the domain $[0,1]$ and range $[0,infty)$ as $f_n(x) = -log frac{nx}{n+1} mathbf{1}_{(0,1]}.$

share|cite|improve this answer

edited Mar 9 at 23:33

answered Mar 9 at 22:02

RRLRRL

52.5k42573

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi RRL. This doesn't seem to address the question. If $f_n^2$ is UC, is $f_n$ UC?
  $endgroup$
  – Mark Viola
  Mar 9 at 22:33
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkViola no, in fact this is on topic, read the title and the second last sentence.
  $endgroup$
  – enedil
  Mar 9 at 22:41
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The wording of the second part is not the best. "I however can not find an example where $f_{n}{underset {n }{rightrightarrows }}f,$ then $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$." If this is not an implication -- i.e. read "and" in place of "then" --then I showed both can hold. "This cannot be true I think, can it?" If by this OP is doubting the implication, then I showed it is false with a counterexample.
  $endgroup$
  – RRL
  Mar 9 at 23:14
  
  
  

  
  
  
  

add a comment | 

$begingroup$

In the first part, the claim $f_{n}{underset {n }{rightrightarrows }}f implies f_{n}^{1/2}{underset {n }{rightrightarrows }}f^{1/2} $ for $f_n,f :[0,1] to [0,infty)$ is true. Your proof starts out correctly, but your last inequality is not valid. You can complete the argument by stating that for all $epsilon > 0$ there exists $N$ independent of $x$ such that when $n > N$ we have for all $x in [0,1]$,

$$|f_n(x) - f(x)| < epsilon^2, $$

and, thus,

$$left|sqrt{f_n(x)} - sqrt{f(x)}right| leqslant sqrt{|f_n(x) - f(x)|} < epsilon.$$

Second Part

Clearly, there are examples where both $f_{n}{underset {n }{rightrightarrows }}f$ and $f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ hold, either by virtue of your first proposition or the trivial example $f_n(x) = 1/n, , f(x) = 0$.

However, the implication$f_{n}{underset {n }{rightrightarrows }}fimplies f^{2}_{n}{underset {n }{rightrightarrows }}f^{2}$ is not true.

For a counterexample, take $f_n,f:(0,1] to mathbb{R}$ where

$$f_n(x) = log frac{nx}{n+1}, quad f(x) = log x$$

We have $f_{n}{underset {n }{rightrightarrows }}f$, since

$$|f_n(x) - f(x)| = log frac{n+1}{n} to 0$$

However,
$$|f_n^2(x) - f^2(x)| = left|log frac{nx}{n+1 } - log xright|left|log frac{nx}{n+1 } + log xright| = logfrac{n+1}{n} left|log frac{nx^2}{n+1 }right|,$$

and by choosing $x_n = 1/(n e^n)$ we have

$$|f_n^2(x_n) - f^2(x_n)| = logfrac{n+1}{n} log (n^2 + n) + 2n logfrac{n+1}{n} $$

Since the RHS does not converge to $0$ as $n to infty$, it follows that $f_{n}^2{underset {n }{rightrightarrows }}f^2$ does not hold on $(0,1].$

This example is easily modified for the domain $[0,1]$ and range $[0,infty)$ as $f_n(x) = -log frac{nx}{n+1} mathbf{1}_{(0,1]}.$

share|cite|improve this answer

edited Mar 9 at 23:33

answered Mar 9 at 22:02

RRLRRL

52.5k42573

$endgroup$

share|cite|improve this answer

edited Mar 9 at 23:33

answered Mar 9 at 22:02

RRLRRL

52.5k42573

answered Mar 9 at 22:02

RRLRRL

52.5k42573

answered Mar 9 at 22:02

RRLRRL

52.5k42573