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Exchange of limits with an integrable singularity

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1

$begingroup$

Consider the following function of the real variable $y$ (it a simplified version of a so-called fidelity in quantum mechanics)

$$F_N(y)=sum_k=1^Nfrac1Nlogleft|y-frackNright|.$$

The mathematical question is: is correct to state that
$$
lim_Nrightarrowinftylim_yto y^*F_N(y)
not = lim_yto y^*lim_Nrightarrow inftyF_N(y),
$$

for any $y^*in[0,1]$?

My naive reasoning is this: if we calculate first the limit $yto y^*$, then increasing $N$ we can find a $x_k=frackN$ arbitrary close to $y^*in[0,1]$, making the logarithm divergent.

If we instead calculate $Nto infty$ first, then $$lim_NrightarrowinftyF_N(y)=-1+(1-y)log(1-y)+ylog(y)$$ and also the limit $yrightarrow y^*$ exists.

Thanks a lot.

real-analysis limits

share|cite|improve this question

edited Mar 28 at 14:53

gt6989b

35.2k22557

asked Mar 28 at 14:51

jacopovitijacopoviti

734

$endgroup$

add a comment | 

1

$begingroup$

Consider the following function of the real variable $y$ (it a simplified version of a so-called fidelity in quantum mechanics)

$$F_N(y)=sum_k=1^Nfrac1Nlogleft|y-frackNright|.$$

The mathematical question is: is correct to state that
$$
lim_Nrightarrowinftylim_yto y^*F_N(y)
not = lim_yto y^*lim_Nrightarrow inftyF_N(y),
$$

for any $y^*in[0,1]$?

My naive reasoning is this: if we calculate first the limit $yto y^*$, then increasing $N$ we can find a $x_k=frackN$ arbitrary close to $y^*in[0,1]$, making the logarithm divergent.

If we instead calculate $Nto infty$ first, then $$lim_NrightarrowinftyF_N(y)=-1+(1-y)log(1-y)+ylog(y)$$ and also the limit $yrightarrow y^*$ exists.

Thanks a lot.

real-analysis limits

share|cite|improve this question

edited Mar 28 at 14:53

gt6989b

35.2k22557

asked Mar 28 at 14:51

jacopovitijacopoviti

734

$endgroup$

add a comment | 

$begingroup$

Consider the following function of the real variable $y$ (it a simplified version of a so-called fidelity in quantum mechanics)

$$F_N(y)=sum_k=1^Nfrac1Nlogleft|y-frackNright|.$$

The mathematical question is: is correct to state that
$$
lim_Nrightarrowinftylim_yto y^*F_N(y)
not = lim_yto y^*lim_Nrightarrow inftyF_N(y),
$$

for any $y^*in[0,1]$?

My naive reasoning is this: if we calculate first the limit $yto y^*$, then increasing $N$ we can find a $x_k=frackN$ arbitrary close to $y^*in[0,1]$, making the logarithm divergent.

If we instead calculate $Nto infty$ first, then $$lim_NrightarrowinftyF_N(y)=-1+(1-y)log(1-y)+ylog(y)$$ and also the limit $yrightarrow y^*$ exists.

Thanks a lot.

real-analysis limits

share|cite|improve this question

edited Mar 28 at 14:53

gt6989b

35.2k22557

asked Mar 28 at 14:51

jacopovitijacopoviti

734

$endgroup$

share|cite|improve this question

edited Mar 28 at 14:53

gt6989b

35.2k22557

asked Mar 28 at 14:51

jacopovitijacopoviti

734

share|cite|improve this question

edited Mar 28 at 14:53

gt6989b

35.2k22557

asked Mar 28 at 14:51

jacopovitijacopoviti

734

edited Mar 28 at 14:53

gt6989b

35.2k22557

edited Mar 28 at 14:53

gt6989b

35.2k22557

asked Mar 28 at 14:51

jacopovitijacopoviti

734

asked Mar 28 at 14:51

jacopovitijacopoviti

734