# Convergence of density estimates with parzen window

I am trying to understand why $$lim_{||u|| \rightarrow+\infty}{\varphi(u)}\prod_{i=1}^{d}u_{i} = 0$$ is necessary for convergence of Parzen density estimates. Similar question has been asked here

Intuitively we estimate density using parzen windows by squeezing a distribution into smaller windows as the number of observations increases. When $$n=\infty$$ this distribution should converge to spike at $$0$$. Thus we design density kernel in a way that probability is 0 everywhere except origin after squeezing infinitely. If the density function has a finite value at someplace close to infinity it will be difficult to squeegee it to a spike at 0. This can be understood in terms of convergence of kernel density function to delta dirac function. I have tried to understand this mathetically as follows:

Convergence in mean square for parzen window assumes that limit of sequence of function in $$\mathbf{R}^m$$ given by $$\epsilon^{-m} \varphi(x/\epsilon)$$ converges to Dirac Delta function due to which we can say $$\bar{p}(x)=\int \frac{1}{V} \varphi(\frac{x-v}{\epsilon}) p(v) dx$$ converges to $$p(x)$$ as $$\epsilon \to 0$$.

Coming to $$lim_{||u|| \rightarrow+\infty}{\varphi(u)}\prod_{i=1}^{d}u_{i} = 0$$.

Substituting $$x_i/\epsilon=u_i$$ for any $$x >0$$ in $$L=lim_{||u|| \rightarrow+\infty}{\varphi(u)}\prod_{i=1}^{d}u_{i}$$ we get

$$L=lim_{||\epsilon|| \rightarrow 0}(\prod_{i=1}^{d}x_{i})\epsilon^{-m} \varphi(x/\epsilon)$$

Since $$lim_{||\epsilon|| \rightarrow 0}\epsilon^{-m} \varphi(x/\epsilon)$$ is delta dirac function its value should be be $$0$$ everywhere except $$x=0$$, so $$L=0$$

Please let me know if my understanding is correct.