Limit of a convolution 2

Question

Let $\phi$ be a non-negative test function with total integral =1 and with support in the ball of radius 1 and centered at the origin in $\mathbb{R}^{n}$. Set $\phi_{t}(x) =t^{-n}\phi(x/t)$. Is the following limit $$\limsup_{t\rightarrow0}\int_{\mathbb{R}^{n}}|\Delta\phi_{t}(y)|dy$$ bounded? ($\Delta$ is the laplacian)

Answer 1

For $t>0$ we can write by definition that $$\int_{\Bbb R^n} |\Delta \phi_t(y)|dy = \int_{\Bbb R^n} t^{-n}| (\partial_{y_1}^2 +\ldots+\partial_{y_n}^2) \phi(y/t)|dy = \int_{\Bbb R^n} t^{-n-2}| (\Delta \phi)(y/t)|dy $$ $$= \int_{\Bbb R^n} t^{-2}| (\Delta \phi)(u)|du. $$

Therefore, unless $\int_{\Bbb R^n}| (\Delta \phi)(u)|du=0$, you limit diverges to $+\infty$. On the other hand, compactly supported harmonic functions are identically zero, therefore for your function $\phi$ you have $\int_{\Bbb R^n}| (\Delta \phi)(u)|du>0$.