$(t_k)_{kgeq 1}$ secuence in $[0,1]$ , $sum a_k <infty$ implies $sum frac{a_k}{sqrt{|t_k-x|}}$ a.e












0














I've read a solution for the following problem but I'm not convinced if solution it's correct.



The problem:
Let $(t_k)_{kgeq 1}$ be a sequence in $[0,1]$ and $(a_k)_{kgeq 1}$ be a sequence of nonnegative real numbers such that $sum_{k=1}^{infty} a_k < infty$. Prove that
$$sum_{k=1}^{infty}frac{a_k}{sqrt{|x-t_k|}}$$
converges for almost all $x in [0,1]$.



The solution:
Consider $f_n(x)=sum_{k=1}^{n} frac{a_k}{sqrt{|x-t_k|}}geq 0 $. Since for each $t in [0,1]$ we have



$$int_{0}^1 frac{dx}{sqrt{|t-x|}}=int_{0}^t frac{dx}{sqrt{t-x}}+int_{t}^1 frac{dx}{sqrt{x-t}}=frac{sqrt{t}}{2}+frac{sqrt{1-t}}{2}leq 1$$
we infer that
$$int_{0}^1 f_n = sum_{k=1}^{n} int_{0}^{1} frac{a_k}{sqrt{|x-t_k|}}leq sum_{k=1}^{n} a_k $$
Hence $(f_n)_n$ is convergent in $L^{1}([0,1])$. In particular by a $textbf{ familiar theorem}$, the numerical sequence $(f_n(x))_n$ will have to converge for almost all $x in [0,1]$



My question is. What familiar theorem? In general convergence in $L^{1}[0,1]$ doesn't imply almost everywhere pointwise convergence. Just imply the existence of a subsequence convergent pointwise a.e.
Please, let me know if that solution is wrong or another aproach to solve it.






real-analysis analysis measure-theory







share|cite|improve this question


















  • 1




    You do see that $f_1 le f_2 le f_3 le dots$, don't you?
    – GEdgar
    Dec 6 at 23:41










  • @GEdgar Yes. Do you mean Monotone Convergence Theorem?
    – Pablo Herrera
    Dec 7 at 2:43
















0














I've read a solution for the following problem but I'm not convinced if solution it's correct.



The problem:
Let $(t_k)_{kgeq 1}$ be a sequence in $[0,1]$ and $(a_k)_{kgeq 1}$ be a sequence of nonnegative real numbers such that $sum_{k=1}^{infty} a_k < infty$. Prove that
$$sum_{k=1}^{infty}frac{a_k}{sqrt{|x-t_k|}}$$
converges for almost all $x in [0,1]$.



The solution:
Consider $f_n(x)=sum_{k=1}^{n} frac{a_k}{sqrt{|x-t_k|}}geq 0 $. Since for each $t in [0,1]$ we have



$$int_{0}^1 frac{dx}{sqrt{|t-x|}}=int_{0}^t frac{dx}{sqrt{t-x}}+int_{t}^1 frac{dx}{sqrt{x-t}}=frac{sqrt{t}}{2}+frac{sqrt{1-t}}{2}leq 1$$
we infer that
$$int_{0}^1 f_n = sum_{k=1}^{n} int_{0}^{1} frac{a_k}{sqrt{|x-t_k|}}leq sum_{k=1}^{n} a_k $$
Hence $(f_n)_n$ is convergent in $L^{1}([0,1])$. In particular by a $textbf{ familiar theorem}$, the numerical sequence $(f_n(x))_n$ will have to converge for almost all $x in [0,1]$



My question is. What familiar theorem? In general convergence in $L^{1}[0,1]$ doesn't imply almost everywhere pointwise convergence. Just imply the existence of a subsequence convergent pointwise a.e.
Please, let me know if that solution is wrong or another aproach to solve it.






real-analysis analysis measure-theory







share|cite|improve this question


















  • 1




    You do see that $f_1 le f_2 le f_3 le dots$, don't you?
    – GEdgar
    Dec 6 at 23:41










  • @GEdgar Yes. Do you mean Monotone Convergence Theorem?
    – Pablo Herrera
    Dec 7 at 2:43














0












0








0







I've read a solution for the following problem but I'm not convinced if solution it's correct.



The problem:
Let $(t_k)_{kgeq 1}$ be a sequence in $[0,1]$ and $(a_k)_{kgeq 1}$ be a sequence of nonnegative real numbers such that $sum_{k=1}^{infty} a_k < infty$. Prove that
$$sum_{k=1}^{infty}frac{a_k}{sqrt{|x-t_k|}}$$
converges for almost all $x in [0,1]$.



The solution:
Consider $f_n(x)=sum_{k=1}^{n} frac{a_k}{sqrt{|x-t_k|}}geq 0 $. Since for each $t in [0,1]$ we have



$$int_{0}^1 frac{dx}{sqrt{|t-x|}}=int_{0}^t frac{dx}{sqrt{t-x}}+int_{t}^1 frac{dx}{sqrt{x-t}}=frac{sqrt{t}}{2}+frac{sqrt{1-t}}{2}leq 1$$
we infer that
$$int_{0}^1 f_n = sum_{k=1}^{n} int_{0}^{1} frac{a_k}{sqrt{|x-t_k|}}leq sum_{k=1}^{n} a_k $$
Hence $(f_n)_n$ is convergent in $L^{1}([0,1])$. In particular by a $textbf{ familiar theorem}$, the numerical sequence $(f_n(x))_n$ will have to converge for almost all $x in [0,1]$



My question is. What familiar theorem? In general convergence in $L^{1}[0,1]$ doesn't imply almost everywhere pointwise convergence. Just imply the existence of a subsequence convergent pointwise a.e.
Please, let me know if that solution is wrong or another aproach to solve it.






real-analysis analysis measure-theory







share|cite|improve this question













I've read a solution for the following problem but I'm not convinced if solution it's correct.



The problem:
Let $(t_k)_{kgeq 1}$ be a sequence in $[0,1]$ and $(a_k)_{kgeq 1}$ be a sequence of nonnegative real numbers such that $sum_{k=1}^{infty} a_k < infty$. Prove that
$$sum_{k=1}^{infty}frac{a_k}{sqrt{|x-t_k|}}$$
converges for almost all $x in [0,1]$.



The solution:
Consider $f_n(x)=sum_{k=1}^{n} frac{a_k}{sqrt{|x-t_k|}}geq 0 $. Since for each $t in [0,1]$ we have



$$int_{0}^1 frac{dx}{sqrt{|t-x|}}=int_{0}^t frac{dx}{sqrt{t-x}}+int_{t}^1 frac{dx}{sqrt{x-t}}=frac{sqrt{t}}{2}+frac{sqrt{1-t}}{2}leq 1$$
we infer that
$$int_{0}^1 f_n = sum_{k=1}^{n} int_{0}^{1} frac{a_k}{sqrt{|x-t_k|}}leq sum_{k=1}^{n} a_k $$
Hence $(f_n)_n$ is convergent in $L^{1}([0,1])$. In particular by a $textbf{ familiar theorem}$, the numerical sequence $(f_n(x))_n$ will have to converge for almost all $x in [0,1]$



My question is. What familiar theorem? In general convergence in $L^{1}[0,1]$ doesn't imply almost everywhere pointwise convergence. Just imply the existence of a subsequence convergent pointwise a.e.
Please, let me know if that solution is wrong or another aproach to solve it.






real-analysis analysis measure-theory




real-analysis analysis measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 at 23:32









Pablo Herrera

397113




397113








  • 1




    You do see that $f_1 le f_2 le f_3 le dots$, don't you?
    – GEdgar
    Dec 6 at 23:41










  • @GEdgar Yes. Do you mean Monotone Convergence Theorem?
    – Pablo Herrera
    Dec 7 at 2:43














  • 1




    You do see that $f_1 le f_2 le f_3 le dots$, don't you?
    – GEdgar
    Dec 6 at 23:41










  • @GEdgar Yes. Do you mean Monotone Convergence Theorem?
    – Pablo Herrera
    Dec 7 at 2:43








1




1




You do see that $f_1 le f_2 le f_3 le dots$, don't you?
– GEdgar
Dec 6 at 23:41




You do see that $f_1 le f_2 le f_3 le dots$, don't you?
– GEdgar
Dec 6 at 23:41












@GEdgar Yes. Do you mean Monotone Convergence Theorem?
– Pablo Herrera
Dec 7 at 2:43




@GEdgar Yes. Do you mean Monotone Convergence Theorem?
– Pablo Herrera
Dec 7 at 2:43










1 Answer
1






active

oldest

votes


















1














First we should note that $f(x):=lim_{n rightarrow infty} f_n(x) = sum_{k=1}^infty a_k |x-t_k|^{-1/2}$ exists in $[0,infty]$. Strictly speaking, this function is not defined in all points $t_k$. But since we have only countable many of them, the corresponding set is a $lambda$-nullset.



Now we can apply the monotone convergence theorem in order to get
$$int_0^1 f(x) , dx = lim_{n rightarrow infty} sum_{k=1}^n a_k int_0^1 |x-t_k|^{-1/2} , dx le sum_{k=1}^infty a_k<infty,$$
where in the last step we have used your argument. Thus $f$ is integrable and this implies that $f(x) < infty$ for almost all $x in [0,1]$ - in other words, the series converges for almost everywhere. In fact, note that
$$C:=int_0^1 f(x) dx ge k lambda{f >k},$$
i.e. $lambda {f >k} le C/k$. Thus letting $k rightarrow infty$, we get $lambda{f= infty} =0$.






share|cite|improve this answer





















  • Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
    – Pablo Herrera
    Dec 7 at 19:26













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1 Answer
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1














First we should note that $f(x):=lim_{n rightarrow infty} f_n(x) = sum_{k=1}^infty a_k |x-t_k|^{-1/2}$ exists in $[0,infty]$. Strictly speaking, this function is not defined in all points $t_k$. But since we have only countable many of them, the corresponding set is a $lambda$-nullset.



Now we can apply the monotone convergence theorem in order to get
$$int_0^1 f(x) , dx = lim_{n rightarrow infty} sum_{k=1}^n a_k int_0^1 |x-t_k|^{-1/2} , dx le sum_{k=1}^infty a_k<infty,$$
where in the last step we have used your argument. Thus $f$ is integrable and this implies that $f(x) < infty$ for almost all $x in [0,1]$ - in other words, the series converges for almost everywhere. In fact, note that
$$C:=int_0^1 f(x) dx ge k lambda{f >k},$$
i.e. $lambda {f >k} le C/k$. Thus letting $k rightarrow infty$, we get $lambda{f= infty} =0$.






share|cite|improve this answer





















  • Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
    – Pablo Herrera
    Dec 7 at 19:26


















1














First we should note that $f(x):=lim_{n rightarrow infty} f_n(x) = sum_{k=1}^infty a_k |x-t_k|^{-1/2}$ exists in $[0,infty]$. Strictly speaking, this function is not defined in all points $t_k$. But since we have only countable many of them, the corresponding set is a $lambda$-nullset.



Now we can apply the monotone convergence theorem in order to get
$$int_0^1 f(x) , dx = lim_{n rightarrow infty} sum_{k=1}^n a_k int_0^1 |x-t_k|^{-1/2} , dx le sum_{k=1}^infty a_k<infty,$$
where in the last step we have used your argument. Thus $f$ is integrable and this implies that $f(x) < infty$ for almost all $x in [0,1]$ - in other words, the series converges for almost everywhere. In fact, note that
$$C:=int_0^1 f(x) dx ge k lambda{f >k},$$
i.e. $lambda {f >k} le C/k$. Thus letting $k rightarrow infty$, we get $lambda{f= infty} =0$.






share|cite|improve this answer





















  • Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
    – Pablo Herrera
    Dec 7 at 19:26
















1












1








1






First we should note that $f(x):=lim_{n rightarrow infty} f_n(x) = sum_{k=1}^infty a_k |x-t_k|^{-1/2}$ exists in $[0,infty]$. Strictly speaking, this function is not defined in all points $t_k$. But since we have only countable many of them, the corresponding set is a $lambda$-nullset.



Now we can apply the monotone convergence theorem in order to get
$$int_0^1 f(x) , dx = lim_{n rightarrow infty} sum_{k=1}^n a_k int_0^1 |x-t_k|^{-1/2} , dx le sum_{k=1}^infty a_k<infty,$$
where in the last step we have used your argument. Thus $f$ is integrable and this implies that $f(x) < infty$ for almost all $x in [0,1]$ - in other words, the series converges for almost everywhere. In fact, note that
$$C:=int_0^1 f(x) dx ge k lambda{f >k},$$
i.e. $lambda {f >k} le C/k$. Thus letting $k rightarrow infty$, we get $lambda{f= infty} =0$.






share|cite|improve this answer












First we should note that $f(x):=lim_{n rightarrow infty} f_n(x) = sum_{k=1}^infty a_k |x-t_k|^{-1/2}$ exists in $[0,infty]$. Strictly speaking, this function is not defined in all points $t_k$. But since we have only countable many of them, the corresponding set is a $lambda$-nullset.



Now we can apply the monotone convergence theorem in order to get
$$int_0^1 f(x) , dx = lim_{n rightarrow infty} sum_{k=1}^n a_k int_0^1 |x-t_k|^{-1/2} , dx le sum_{k=1}^infty a_k<infty,$$
where in the last step we have used your argument. Thus $f$ is integrable and this implies that $f(x) < infty$ for almost all $x in [0,1]$ - in other words, the series converges for almost everywhere. In fact, note that
$$C:=int_0^1 f(x) dx ge k lambda{f >k},$$
i.e. $lambda {f >k} le C/k$. Thus letting $k rightarrow infty$, we get $lambda{f= infty} =0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 at 12:38









p4sch

4,800217




4,800217












  • Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
    – Pablo Herrera
    Dec 7 at 19:26




















  • Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
    – Pablo Herrera
    Dec 7 at 19:26


















Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
– Pablo Herrera
Dec 7 at 19:26






Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that?
– Pablo Herrera
Dec 7 at 19:26




















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