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edited Jul 16 at 15:49

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I don't want to come across as unnecessarily pedantic without being helpful, so I'm guessing what the question really means. Under my guessed assumptions, the answer is in general no even for $f_n = f_0 = 0$ for all $n \in \mathbb{N}$.

For $n \in \mathbb{N}$, let $\beta_{0, 1, n} = 0$, $\beta_{0, 2, n} = n$, $\beta_{1, k, n} = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_{1, n} = 1 - \frac{1}{n}$, $\omega_{2, n} = 1 - \frac{1}{n}$$\omega_{2, n} = \frac{1}{n}$ and $\omega_{k, n} = 0$ for $k \geq 2$.

Regarding the limit, let $\beta_{0, 1}^* = 0$ and $\beta_{1, k}^* = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_1^* = 1$.

Then $$ \int \, \nu(\mathrm{d}x) \, \Vert p_n(\cdot \vert x) - p_0(\cdot \vert x) \Vert_{L^1} = \int \, \nu(\mathrm{d}x) \, \int_{\mathbb{R}} \, \mathrm{d}y \, \vert p_n(y \vert x) - p_0(y \vert x)\vert \leq \frac{2}{n} \xrightarrow[n \to \infty]{} 0 $$ but $$ \int \, \nu(\mathrm{d}x) \, \left\vert \omega_{1, n} \beta_{0, 1, n} + \omega_{2, n} \beta_{0, 2, n} - \omega_{1}^* \beta_{0, 1}^* \right\vert = \int \, \nu(\mathrm{d}x) \, \left\vert 0 + n \frac{1}{n} - 0\right\vert = 1, $$ hence the desired convergence does not hold.

I don't want to come across as unnecessarily pedantic without being helpful, so I'm guessing what the question really means. Under my guessed assumptions, the answer is in general no even for $f_n = f_0 = 0$ for all $n \in \mathbb{N}$.

For $n \in \mathbb{N}$, let $\beta_{0, 1, n} = 0$, $\beta_{0, 2, n} = n$, $\beta_{1, k, n} = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_{1, n} = 1 - \frac{1}{n}$, $\omega_{2, n} = 1 - \frac{1}{n}$ and $\omega_{k, n} = 0$ for $k \geq 2$.

Regarding the limit, let $\beta_{0, 1}^* = 0$ and $\beta_{1, k}^* = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_1^* = 1$.

Then $$ \int \, \nu(\mathrm{d}x) \, \Vert p_n(\cdot \vert x) - p_0(\cdot \vert x) \Vert_{L^1} = \int \, \nu(\mathrm{d}x) \, \int_{\mathbb{R}} \, \mathrm{d}y \, \vert p_n(y \vert x) - p_0(y \vert x)\vert \leq \frac{2}{n} \xrightarrow[n \to \infty]{} 0 $$ but $$ \int \, \nu(\mathrm{d}x) \, \left\vert \omega_{1, n} \beta_{0, 1, n} + \omega_{2, n} \beta_{0, 2, n} - \omega_{1}^* \beta_{0, 1}^* \right\vert = \int \, \nu(\mathrm{d}x) \, \left\vert 0 + n \frac{1}{n} - 0\right\vert = 1, $$ hence the desired convergence does not hold.

I don't want to come across as unnecessarily pedantic without being helpful, so I'm guessing what the question really means. Under my guessed assumptions, the answer is in general no even for $f_n = f_0 = 0$ for all $n \in \mathbb{N}$.

For $n \in \mathbb{N}$, let $\beta_{0, 1, n} = 0$, $\beta_{0, 2, n} = n$, $\beta_{1, k, n} = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_{1, n} = 1 - \frac{1}{n}$, $\omega_{2, n} = \frac{1}{n}$ and $\omega_{k, n} = 0$ for $k \geq 2$.

Regarding the limit, let $\beta_{0, 1}^* = 0$ and $\beta_{1, k}^* = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_1^* = 1$.

Then $$ \int \, \nu(\mathrm{d}x) \, \Vert p_n(\cdot \vert x) - p_0(\cdot \vert x) \Vert_{L^1} = \int \, \nu(\mathrm{d}x) \, \int_{\mathbb{R}} \, \mathrm{d}y \, \vert p_n(y \vert x) - p_0(y \vert x)\vert \leq \frac{2}{n} \xrightarrow[n \to \infty]{} 0 $$ but $$ \int \, \nu(\mathrm{d}x) \, \left\vert \omega_{1, n} \beta_{0, 1, n} + \omega_{2, n} \beta_{0, 2, n} - \omega_{1}^* \beta_{0, 1}^* \right\vert = \int \, \nu(\mathrm{d}x) \, \left\vert 0 + n \frac{1}{n} - 0\right\vert = 1, $$ hence the desired convergence does not hold.

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created Jul 16 at 14:48

578
1
2
9

I don't want to come across as unnecessarily pedantic without being helpful, so I'm guessing what the question really means. Under my guessed assumptions, the answer is in general no even for $f_n = f_0 = 0$ for all $n \in \mathbb{N}$.

For $n \in \mathbb{N}$, let $\beta_{0, 1, n} = 0$, $\beta_{0, 2, n} = n$, $\beta_{1, k, n} = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_{1, n} = 1 - \frac{1}{n}$, $\omega_{2, n} = 1 - \frac{1}{n}$ and $\omega_{k, n} = 0$ for $k \geq 2$.

Regarding the limit, let $\beta_{0, 1}^* = 0$ and $\beta_{1, k}^* = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_1^* = 1$.

Then $$ \int \, \nu(\mathrm{d}x) \, \Vert p_n(\cdot \vert x) - p_0(\cdot \vert x) \Vert_{L^1} = \int \, \nu(\mathrm{d}x) \, \int_{\mathbb{R}} \, \mathrm{d}y \, \vert p_n(y \vert x) - p_0(y \vert x)\vert \leq \frac{2}{n} \xrightarrow[n \to \infty]{} 0 $$ but $$ \int \, \nu(\mathrm{d}x) \, \left\vert \omega_{1, n} \beta_{0, 1, n} + \omega_{2, n} \beta_{0, 2, n} - \omega_{1}^* \beta_{0, 1}^* \right\vert = \int \, \nu(\mathrm{d}x) \, \left\vert 0 + n \frac{1}{n} - 0\right\vert = 1, $$ hence the desired convergence does not hold.