Let $f\in C(\mathbb{R}^{N}); supp(f) \subset \{ x:||x||<1 \}; \int_{\mathbb{R}^{N}} f(x)dx = 1$ and consider the sequence $f_{k}=k^{N}f(kx)$. I need to show that $f_{k}(x) \to \delta(x)$ as $k \to \infty$ in $\mathcal{D'}(\mathbb{R}^{N})$.
To be honest, I'm not completely sure my approach is correct, but I've taken a function $g \in \mathcal{D}(\mathbb{R}^{N})$ and tried to get to the result by manipulating the following integral:
$$\int_{\mathbb{R}^{N}}f_{k}(x)g(x)dx$$
After adding and subtracting a $g(0)$ inside the integral (so I could use the mean value theorem), using a few inequalities and changing the variable n times, I got to the following inequality:
$$\int_{\mathbb{R}^{N}}f_{k}(x)g(x)dx \le C + g(0)$$
where $C$ is a constant. (which I got from $max_{x \in \mathbb{R}^{N}}|\nabla g|$)
I might have gone wrong somewhere along the way; if I didn't, I'm not sure how to go on. Any advice or hints would be greatly appreciated. (I do not necessarily need a complete solution as I'd rather do it myself)
Hint: $\int f_k(x)g(x)\, dx = \int f(x)g(x/k)\, dx.$