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I am dealing with the following limit involving a sequence of ditributions:

$$\lim_{n \to \infty}T_n= \lim_{n \to \infty}\frac{1}{n^2}\sum_{k=1}^{2n}k \delta_{\frac kn} $$ Using the definition of Delta distribution I find that: $$ \lim_{n \to \infty} <T_n,\phi> = \lim_{n \to \infty} \frac{1}{n^2}\sum_{k=1}^{2n}k \phi\left(\frac kn\right)$$ and here is where I get stuck. I was trying to "see" some integral sum there... Any idea how to proceed?

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Write $$\frac{1}{n^2}\sum_{k=1}^{2n}k \phi\left(\frac kn\right) = \frac{1}{n}\sum_{k=1}^{2n} \frac kn \phi\left(\frac kn\right)$$

and we can think of

$$\frac{1}{n}\sum_{k=1}^{2n} \frac kn \phi\left(\frac kn\right)$$

as the Riemann sum of the function $x\phi (x)$ over the interval $[0,2]$. Thus

$$\lim_{n\to \infty } \frac{1}{n^2}\sum_{k=1}^{2n}k\phi\left(\frac kn\right) = \int_0^2 x\phi(x) \mathrm d x. $$

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  • $\begingroup$ Thanks! So it seems that I can conclude that $T_n \to T_f$ where $f=\frac x2 \mathbb{1}_{[0,2]}$ right? $\endgroup$ – Leonardo Massai May 22 '18 at 11:33
  • $\begingroup$ Yes, @LeonardoMassai $\endgroup$ – user99914 May 22 '18 at 15:32
  • $\begingroup$ Hi, I have updated the answer, there shouldn't be a 1/2. @LeonardoMassai $\endgroup$ – user99914 May 25 '18 at 4:37
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Hint $$\frac{1}{n^2}\sum_{k=1}^{2n}k \phi\left(\frac kn\right)=\frac{1}{n}\sum_{k=1}^{2n}\frac{k}{n} \phi\left(\frac kn\right)=\frac{1}{n}\sum_{k=1}^{2n} \psi\left(\frac kn\right)$$ where $\psi(x)=x\phi(x)$ so it is an integral sum.

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