Uniform convergence of $f_n(x) = sqrt{x^2 + frac{1}{n^2}}$, solution verification












0












$begingroup$


Is my reasoning right? I have $f_n(x) = sqrt{x^2 + frac{1}{n^2}}$ for $x in mathbb{R}$, so I conclude that it's pointwise convergent $f_n to |x|$, and moreover it's uniformly convergent to $|x|$, because $left | sqrt{x^2 + frac{1}{n^2}}- |x| right | = frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2} to 0$



That's because $sqrt{x^2 + frac{1}{n^2}} to |x|$ and it's decreasing for any $x in mathbb{R}$, therefore I can make denominator smaller by $sqrt{x^2 + frac{1}{n^2}} leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{1}{n^2}$, otherwise if $|x| leq frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{2|x|}{n^22|x|} = frac{1}{n^2}$






proof-verification uniform-convergence sequence-of-function







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$endgroup$












  • $begingroup$
    Yes, correct. $quad quad quad$
    $endgroup$
    – Rebellos
    Dec 18 '18 at 21:21






  • 1




    $begingroup$
    There is a mistake, note that $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n}$$ That is: you want to make the denominator as small as possible to bound the fraction above. The maximum happen when $x=0$, hence the above inequality
    $endgroup$
    – Masacroso
    Dec 18 '18 at 21:23








  • 2




    $begingroup$
    The bounds $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2}$$ and, if $|x| leq frac{1}{2}$, $$frac{1}{n^22|x|} leq frac{1}{n^2}$$ both fail to hold.
    $endgroup$
    – Did
    Dec 18 '18 at 21:25










  • $begingroup$
    @Did Right, when $|x| leq frac{1}{2}$, then $2|x| leq 1$, so $frac{1}{n^2 2|x|} = frac{1}{n} cdot frac{1}{n2|x|} leq frac{1}{n} to 0$, because $n2|x| geq 1$ for fixed $n$ and for every $|x| leq frac{1}{2}$ without $0$, we can check as a special case that for $x=0$ boundary holds too, daym I did it too fast and got that mistake
    $endgroup$
    – chandx
    Dec 18 '18 at 22:11










  • $begingroup$
    "because n2|x|≥1 for fixed n and for every |x|≤12 without 0" ?? Well, no.
    $endgroup$
    – Did
    Dec 18 '18 at 22:15
















0












$begingroup$


Is my reasoning right? I have $f_n(x) = sqrt{x^2 + frac{1}{n^2}}$ for $x in mathbb{R}$, so I conclude that it's pointwise convergent $f_n to |x|$, and moreover it's uniformly convergent to $|x|$, because $left | sqrt{x^2 + frac{1}{n^2}}- |x| right | = frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2} to 0$



That's because $sqrt{x^2 + frac{1}{n^2}} to |x|$ and it's decreasing for any $x in mathbb{R}$, therefore I can make denominator smaller by $sqrt{x^2 + frac{1}{n^2}} leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{1}{n^2}$, otherwise if $|x| leq frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{2|x|}{n^22|x|} = frac{1}{n^2}$






proof-verification uniform-convergence sequence-of-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, correct. $quad quad quad$
    $endgroup$
    – Rebellos
    Dec 18 '18 at 21:21






  • 1




    $begingroup$
    There is a mistake, note that $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n}$$ That is: you want to make the denominator as small as possible to bound the fraction above. The maximum happen when $x=0$, hence the above inequality
    $endgroup$
    – Masacroso
    Dec 18 '18 at 21:23








  • 2




    $begingroup$
    The bounds $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2}$$ and, if $|x| leq frac{1}{2}$, $$frac{1}{n^22|x|} leq frac{1}{n^2}$$ both fail to hold.
    $endgroup$
    – Did
    Dec 18 '18 at 21:25










  • $begingroup$
    @Did Right, when $|x| leq frac{1}{2}$, then $2|x| leq 1$, so $frac{1}{n^2 2|x|} = frac{1}{n} cdot frac{1}{n2|x|} leq frac{1}{n} to 0$, because $n2|x| geq 1$ for fixed $n$ and for every $|x| leq frac{1}{2}$ without $0$, we can check as a special case that for $x=0$ boundary holds too, daym I did it too fast and got that mistake
    $endgroup$
    – chandx
    Dec 18 '18 at 22:11










  • $begingroup$
    "because n2|x|≥1 for fixed n and for every |x|≤12 without 0" ?? Well, no.
    $endgroup$
    – Did
    Dec 18 '18 at 22:15














0












0








0





$begingroup$


Is my reasoning right? I have $f_n(x) = sqrt{x^2 + frac{1}{n^2}}$ for $x in mathbb{R}$, so I conclude that it's pointwise convergent $f_n to |x|$, and moreover it's uniformly convergent to $|x|$, because $left | sqrt{x^2 + frac{1}{n^2}}- |x| right | = frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2} to 0$



That's because $sqrt{x^2 + frac{1}{n^2}} to |x|$ and it's decreasing for any $x in mathbb{R}$, therefore I can make denominator smaller by $sqrt{x^2 + frac{1}{n^2}} leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{1}{n^2}$, otherwise if $|x| leq frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{2|x|}{n^22|x|} = frac{1}{n^2}$






proof-verification uniform-convergence sequence-of-function







share|cite|improve this question









$endgroup$




Is my reasoning right? I have $f_n(x) = sqrt{x^2 + frac{1}{n^2}}$ for $x in mathbb{R}$, so I conclude that it's pointwise convergent $f_n to |x|$, and moreover it's uniformly convergent to $|x|$, because $left | sqrt{x^2 + frac{1}{n^2}}- |x| right | = frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2} to 0$



That's because $sqrt{x^2 + frac{1}{n^2}} to |x|$ and it's decreasing for any $x in mathbb{R}$, therefore I can make denominator smaller by $sqrt{x^2 + frac{1}{n^2}} leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{1}{n^2}$, otherwise if $|x| leq frac{1}{2}$ then $frac{1}{n^22|x|} leq frac{2|x|}{n^22|x|} = frac{1}{n^2}$






proof-verification uniform-convergence sequence-of-function




proof-verification uniform-convergence sequence-of-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 21:16









chandxchandx

628




628












  • $begingroup$
    Yes, correct. $quad quad quad$
    $endgroup$
    – Rebellos
    Dec 18 '18 at 21:21






  • 1




    $begingroup$
    There is a mistake, note that $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n}$$ That is: you want to make the denominator as small as possible to bound the fraction above. The maximum happen when $x=0$, hence the above inequality
    $endgroup$
    – Masacroso
    Dec 18 '18 at 21:23








  • 2




    $begingroup$
    The bounds $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2}$$ and, if $|x| leq frac{1}{2}$, $$frac{1}{n^22|x|} leq frac{1}{n^2}$$ both fail to hold.
    $endgroup$
    – Did
    Dec 18 '18 at 21:25










  • $begingroup$
    @Did Right, when $|x| leq frac{1}{2}$, then $2|x| leq 1$, so $frac{1}{n^2 2|x|} = frac{1}{n} cdot frac{1}{n2|x|} leq frac{1}{n} to 0$, because $n2|x| geq 1$ for fixed $n$ and for every $|x| leq frac{1}{2}$ without $0$, we can check as a special case that for $x=0$ boundary holds too, daym I did it too fast and got that mistake
    $endgroup$
    – chandx
    Dec 18 '18 at 22:11










  • $begingroup$
    "because n2|x|≥1 for fixed n and for every |x|≤12 without 0" ?? Well, no.
    $endgroup$
    – Did
    Dec 18 '18 at 22:15


















  • $begingroup$
    Yes, correct. $quad quad quad$
    $endgroup$
    – Rebellos
    Dec 18 '18 at 21:21






  • 1




    $begingroup$
    There is a mistake, note that $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n}$$ That is: you want to make the denominator as small as possible to bound the fraction above. The maximum happen when $x=0$, hence the above inequality
    $endgroup$
    – Masacroso
    Dec 18 '18 at 21:23








  • 2




    $begingroup$
    The bounds $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2}$$ and, if $|x| leq frac{1}{2}$, $$frac{1}{n^22|x|} leq frac{1}{n^2}$$ both fail to hold.
    $endgroup$
    – Did
    Dec 18 '18 at 21:25










  • $begingroup$
    @Did Right, when $|x| leq frac{1}{2}$, then $2|x| leq 1$, so $frac{1}{n^2 2|x|} = frac{1}{n} cdot frac{1}{n2|x|} leq frac{1}{n} to 0$, because $n2|x| geq 1$ for fixed $n$ and for every $|x| leq frac{1}{2}$ without $0$, we can check as a special case that for $x=0$ boundary holds too, daym I did it too fast and got that mistake
    $endgroup$
    – chandx
    Dec 18 '18 at 22:11










  • $begingroup$
    "because n2|x|≥1 for fixed n and for every |x|≤12 without 0" ?? Well, no.
    $endgroup$
    – Did
    Dec 18 '18 at 22:15
















$begingroup$
Yes, correct. $quad quad quad$
$endgroup$
– Rebellos
Dec 18 '18 at 21:21




$begingroup$
Yes, correct. $quad quad quad$
$endgroup$
– Rebellos
Dec 18 '18 at 21:21




1




1




$begingroup$
There is a mistake, note that $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n}$$ That is: you want to make the denominator as small as possible to bound the fraction above. The maximum happen when $x=0$, hence the above inequality
$endgroup$
– Masacroso
Dec 18 '18 at 21:23






$begingroup$
There is a mistake, note that $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n}$$ That is: you want to make the denominator as small as possible to bound the fraction above. The maximum happen when $x=0$, hence the above inequality
$endgroup$
– Masacroso
Dec 18 '18 at 21:23






2




2




$begingroup$
The bounds $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2}$$ and, if $|x| leq frac{1}{2}$, $$frac{1}{n^22|x|} leq frac{1}{n^2}$$ both fail to hold.
$endgroup$
– Did
Dec 18 '18 at 21:25




$begingroup$
The bounds $$frac{1}{n^2 left ( sqrt{x^2 + frac{1}{n^2}} + |x| right )} leq frac{1}{n^2}$$ and, if $|x| leq frac{1}{2}$, $$frac{1}{n^22|x|} leq frac{1}{n^2}$$ both fail to hold.
$endgroup$
– Did
Dec 18 '18 at 21:25












$begingroup$
@Did Right, when $|x| leq frac{1}{2}$, then $2|x| leq 1$, so $frac{1}{n^2 2|x|} = frac{1}{n} cdot frac{1}{n2|x|} leq frac{1}{n} to 0$, because $n2|x| geq 1$ for fixed $n$ and for every $|x| leq frac{1}{2}$ without $0$, we can check as a special case that for $x=0$ boundary holds too, daym I did it too fast and got that mistake
$endgroup$
– chandx
Dec 18 '18 at 22:11




$begingroup$
@Did Right, when $|x| leq frac{1}{2}$, then $2|x| leq 1$, so $frac{1}{n^2 2|x|} = frac{1}{n} cdot frac{1}{n2|x|} leq frac{1}{n} to 0$, because $n2|x| geq 1$ for fixed $n$ and for every $|x| leq frac{1}{2}$ without $0$, we can check as a special case that for $x=0$ boundary holds too, daym I did it too fast and got that mistake
$endgroup$
– chandx
Dec 18 '18 at 22:11












$begingroup$
"because n2|x|≥1 for fixed n and for every |x|≤12 without 0" ?? Well, no.
$endgroup$
– Did
Dec 18 '18 at 22:15




$begingroup$
"because n2|x|≥1 for fixed n and for every |x|≤12 without 0" ?? Well, no.
$endgroup$
– Did
Dec 18 '18 at 22:15










1 Answer
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$begingroup$

A simpler way to proceed: If $a,bge 0,$ then $sqrt {a+b}-sqrt a le sqrt b.$ Proof: Move $sqrt a$ to the other side and square. It follows that



$$sqrt {x^2+1/n^2} -sqrt {x^2} le sqrt {1/n^2} = 1/n.$$






share|cite|improve this answer









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    1 Answer
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    active

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    $begingroup$

    A simpler way to proceed: If $a,bge 0,$ then $sqrt {a+b}-sqrt a le sqrt b.$ Proof: Move $sqrt a$ to the other side and square. It follows that



    $$sqrt {x^2+1/n^2} -sqrt {x^2} le sqrt {1/n^2} = 1/n.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      A simpler way to proceed: If $a,bge 0,$ then $sqrt {a+b}-sqrt a le sqrt b.$ Proof: Move $sqrt a$ to the other side and square. It follows that



      $$sqrt {x^2+1/n^2} -sqrt {x^2} le sqrt {1/n^2} = 1/n.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        A simpler way to proceed: If $a,bge 0,$ then $sqrt {a+b}-sqrt a le sqrt b.$ Proof: Move $sqrt a$ to the other side and square. It follows that



        $$sqrt {x^2+1/n^2} -sqrt {x^2} le sqrt {1/n^2} = 1/n.$$






        share|cite|improve this answer









        $endgroup$



        A simpler way to proceed: If $a,bge 0,$ then $sqrt {a+b}-sqrt a le sqrt b.$ Proof: Move $sqrt a$ to the other side and square. It follows that



        $$sqrt {x^2+1/n^2} -sqrt {x^2} le sqrt {1/n^2} = 1/n.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 21:47









        zhw.zhw.

        74.3k43175




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