Suppose $g$ is the triangle function on $[-1,1]$ and $g(0) = 1$, is it true that $$pg^p(x) \xrightarrow{\mathcal{D}^*} 2 \delta_0?$$
My calculation, we know $\int_\mathbb R g^p(x) dx = \frac{2}{p+1}$ $$\int_\mathbb R p g^p(x) \phi(0) dx = \frac{2p}{p+1} \phi(0) \quad \longrightarrow \quad 2\phi(0).$$
First only look at the right side of $g$, we have $g(x) = 1-x$ on $[0,1]$. It suffices to show that $$\int_0^\infty p(1-x)^p |\phi(x) - \phi(0)| dx $$ tends to zero. Assume $\phi(0) = 0$, we have $$\int_0^\delta p(1-x)^p |\phi(x)| dx + \int_\delta^\infty p(1-x)^p |\phi(x)| dx.$$ The first integral is bounded by $C\max_{[0,\delta]} \phi$; the second integral tends to zero as $p$ tends to infinity for each $\delta >0$. So for each $\epsilon >0$, we can first choose $\delta$ such that the first integral is less than $\epsilon/2$, and take $p$ large will make the second integral less than $\epsilon/2$.
Yes, it's true and your proof is correct. Here is a different proof: let $F_p(x) = \int_{-\infty}^x pg^p(t)\,dt$. Then $$ F_p(x) = \begin{cases} 0, \quad & x\le -1 \\ \frac{p}{p+1}(1+x)^{p+1},\quad & -1\le x\le 0 \\ \frac{2p}{p+1} - \frac{p}{p+1}(1-x)^{p+1},\quad & 0\le x\le 1 \\ \frac{2p}{p+1} , \quad & x\ge 1 \\ \end{cases}$$ As $p\to\infty$, this function converges to $2H$, where $H$ is the Heaviside function. Convergence is pointwise, and everything it bounded, so by the dominated convergence theorem we have $F_p\to 2H$ in $L_1([-A,A])$ for any $A$. Consequently, $F_p\to 2H$ in the sense of distributions. Distributional convergence passes to derivatives, so $F_p'\to 2\delta_0$ as claimed.