Given the space $F$ defined by: $$F=\left\{f\in C^2(\mathbb{R}_+^2;\mathbb{R}):f(x,0)=\int_\mathbb{R} f(z,x)g(z)dz, x>0\right\},$$ I want to prove that the subspace $E$ of $F$ defined by $E=\bigcap_{n\geq 1} E_n$ where $$E_n=\left\{f\in C^2(\mathbb{R}_+^2;\mathbb{R}):f(x,0)=\sum_{k\in\mathbb{N}} f\left(\frac{k}{n},x\right)g_n(k), x>0\right\}$$ is dense in $F$, assuming that $\displaystyle \lim_{n\rightarrow\infty}\sum_{k\in\mathbb{N}} f\left(\frac{k}{n},x\right)g_n(k)=\int_\mathbb{R} f(z,x)g(z)dz$ for all $f$ in $C^2(\mathbb{R}_+^2;\mathbb{R})$, and where $g_n$ and $g$ are positively supported probability distribution functions such that $\frac{X_n}{n}$ converges in distribution to $X$ if $X_n\sim g_n$ and $X\sim g$.

The problem essentially comes from the interchange between the first and second components in the nonlocal boundary condition. More precisely, had the problem been to show that $G=\bigcap_{n\geq 1} G_n$ where: $$G_n=\left\{f\in C^2(\mathbb{R}_+^2;\mathbb{R}):f(x,0)=\sum_{k\in\mathbb{N}} f\left(x,\frac{k}{n}\right)g_n(k), x>0\right\}$$ were dense in: $$H=\left\{f\in C^2(\mathbb{R}_+^2;\mathbb{R}):f(x,0)=\int_\mathbb{R} f(x,z)g(z)dz, x>0\right\},$$ it would have sufficed to define the sequence of functions $f_n$ by: $$ f_n(x,y)=f(x,y)+\frac{y}{\sum_{j\in\mathbb{N}}\frac{j}{n}g_n(j)}\left(\int_\mathbb{R} f(x,z)g(z)dz-\sum_{j\in\mathbb{N}} f\left(x,\frac{j}{n}\right)g_n(j)\right)$$ to deduce that $f_n(x,0)=f(x,0)$, $\displaystyle \sum_{k\in\mathbb{N}} f_n\left(x,\frac{k}{n}\right)g_n(k)=\int_\mathbb{R} f(x,z)g(z)dz$ and $\lim_{n\rightarrow\infty}\|f_n-f\|=0$, obviously with the inverted assumption that $\displaystyle \lim_{n\rightarrow\infty}\sum_{k\in\mathbb{N}} f\left(x,\frac{k}{n}\right)g_n(k)=\int_\mathbb{R} f(x,z)g(z)dz$.

Unfortunately, I have not been able to find a suitable sequence of functions to use in my initial problem where the components are interchanged in the boundary condition. Any ideas or references to literature would be greatly appreciated.

Update: As an example, the sequence of functions $f_n$ defined by: $$ f_n(x,y)=f(x,y)+\mathbb{1}_{\{y>0\}}\left(\int_\mathbb{R} f(z,y)g(z)dz-\sum_{j\in\mathbb{N}} f\left(\frac{j}{n},y\right)g_n(j)\right)$$ would verify $f_n(x,0)=f(x,0)$, $\displaystyle \sum_{k\in\mathbb{N}} f_n\left(\frac{k}{n},x\right)g_n(k)=\int_\mathbb{R} f(z,x)g(z)dz$ and $\lim_{n\rightarrow\infty}\|f_n-f\|=0$, but $f_n$ is discontinuous (I am looking for a function in $C^2(\mathbb{R}_+^2;\mathbb{R}))$.

  • $\begingroup$ Does anyone have an idea how to do this? $\endgroup$
    – user223935
    Commented Aug 24, 2015 at 4:11
  • $\begingroup$ If $f \in C(\Bbb{R}^2_+, \Bbb{R})$ how can you consider $$\int_\mathbb{R} f(z,x)g(z)dz$$ shouldn't it be $$\int_{\mathbb{R}_+} f(z,x)g(z)dz \quad?$$ $\endgroup$ Commented Aug 24, 2015 at 14:17
  • $\begingroup$ $F = F_g$? $E_n = E_{n,g_n}$? how do we choose $g_n$ is it a function of $g$? $\endgroup$ Commented Aug 24, 2015 at 14:34
  • $\begingroup$ @ConradoCosta I forgot to mention, the two integrals you wrote are equivalent since $g$ is a probability distribution function with positive support $\endgroup$
    – user223935
    Commented Aug 24, 2015 at 15:58
  • $\begingroup$ What about $g_n$? is it related to $g$? in which way? $\endgroup$ Commented Aug 24, 2015 at 16:01

1 Answer 1


Consider $$ f_n(x,y)=f(x,y)+ \frac{h(y)}{\sum_k h(k/n) g_n(k)}\left(\int_\mathbb{R} f(z,y)g(z)dz-\sum_{j\in\mathbb{N}} f\left(\frac{j}{n},y\right)g_n(j)\right)$$

with $h_n(y) = ne^{-1/y}$ if $y>0$ and $h_n(y) = 0$ if $y \leq 0$

then it verifies:

$$\sum_k f_n(x,k/n) g_n(k)=\sum f_n(x,k/n) g_n(k)+ \sum_k\frac{h_n(y/n)g_n(k)}{\sum_k h_n(k/n) g_n(k)}\left(\int_\mathbb{R} f(z,y)g(z)dz-\sum_{j\in\mathbb{N}} f\left(\frac{j}{n},y\right)g_n(j)\right) = \int_\mathbb{R} f(z,y)g(z)dz $$

$$\|f_n -f \| \to 0$$

and $f_n\in C^2(\Bbb{R}^2_+, \Bbb{R})$

  • $\begingroup$ thanks for your help man, but I need to find a function $f_n$ which satisfies $f_n(x,0)=\sum_{k\in\mathbb{N}} f_n\left(\frac{k}{n},x\right)g_n(k)$ (and of course with $\lim_{n\rightarrow\infty}\|f_n-f\|=0$. Do you know how to do this? $\endgroup$
    – user223935
    Commented Aug 25, 2015 at 15:55
  • $\begingroup$ In fact the main difficulty I believe comes from the interchange between the variables in the boundary condition, as opposed to the easy case (density of $G$ in $H$) I presented my question $\endgroup$
    – user223935
    Commented Aug 25, 2015 at 15:56
  • $\begingroup$ I will think about it, by the way I don't see how your example $f_n(x,y)=f(x,y)+\frac{y}{\sum_{j\in\mathbb{N}}\frac{j}{n}g_n(j)}\left(\int_R f(x,z)g(z)dz-\sum_{j\in\mathbb{N}} f\left(x,\frac{j}{n}\right)g_n(j)\right)$ satisfies $f_n(x,0)=\sum_{k\in\mathbb{N}} f_n\left(\frac{k}{n},x\right)g_n(k)$. Maybe that's why I forgot this condition. $\endgroup$ Commented Aug 25, 2015 at 15:59
  • $\begingroup$ no this is the easy case (density of $G$ in $H$) where $f_n$ satisfies $f_n(x,0)=f(x,0)$, $\displaystyle \sum_{k\in\mathbb{N}} f_n\left(x,\frac{k}{n},\right)g_n(k)=\int_\mathbb{R} f(x,z)g(z)dz$ (no interchange of variables), but I want to prove density of $E$ in $F$ where there is an interchange of variables, so I need $f_n(x,0)=f(x,0)$, $\displaystyle \sum_{k\in\mathbb{N}} f_n\left(\frac{k}{n},x\right)g_n(k)=\int_\mathbb{R} f(z,x)g(z)dz$. $\endgroup$
    – user223935
    Commented Aug 25, 2015 at 16:07
  • $\begingroup$ I found one function which could work $f_n(x,y)=f(x,y)+\mathbb{1}_{\{y>0\}}\left(\int_\mathbb{R} f(z,y)g(z)dz-\sum_{j\in\mathbb{N}} f\left(\frac{j}{n},y\right)g_n(j)\right)$ but it is not even continuous! anyway thanks for your help $\endgroup$
    – user223935
    Commented Aug 25, 2015 at 16:08

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