# Bounding a gradient of a function

Defining an infinitely differentiable fucntion $\phi$ as $$\phi(x) = \left(\frac d2 \right)^{-n} \int_{\mathbb R^n} \psi\left( \frac{y-x}{d/2} \right) \, dy,$$ I need to show that $$|\text{grad} \, \phi(x)| \le \frac kd,$$ where $k$ is constant, provided that $\phi(x)=0$ for $|x| \ge 1$ and $$\int_{\mathbb R^n} \psi(x) \, dx = 1.$$

My first thoughts are to move the gradient $\text{grad}=\nabla$ inside the integral, like this: $$\nabla \phi(x) = \left(\frac d2 \right)^{-n} \int_{\mathbb R^n} \nabla\psi\left( \frac{y-x}{d/2} \right) \, dy$$ Is this a plausible first step? If so, I thought about taking the gradient of $\phi(\frac{y-x}{d/2})$ after that.

The task seems somewhat ambiguous to me. The way I see it. $$\int_{\mathbb{R}^n}\psi\left(\frac{y-x}{d/2}\right)dy=\left(\frac{d}{2}\right)^n\int_{\mathbb{R}^n}\psi(y-x)dy,\tag 1$$ if the function is expanded by the factor of $d/2$ in all of $n$ directions (this is what $y-x\to\frac{y-x}{d/2}$ does), its integral is multiplied by a factor $(d/2)^n$. $$\int_{\mathbb{R}^n}\psi(y-x)dy=\int_{\mathbb{R}^n}\psi(y)dy, \tag 2$$ the constant shift by $x$ does not change the integral by the whole $\mathbb{R}^n$.
Substituting $(1)$ and $(2)$ to the definition of $\phi(x)$, together with $\int_{\mathbb{R}^n}\psi(y)dy=1$, we get: $$\phi(x)=\left(\frac{d}{2}\right)^{-n}\int_{\mathbb{R}^n}\psi\left(\frac{y-x}{d/2}\right)dy=\left(\frac{d}{2}\right)^{-n}\left(\frac{d}{2}\right)^n\int_{\mathbb{R}^n}\psi(y-x)dy=\\=\int_{\mathbb{R}^n}\psi(y-x)dy=\int_{\mathbb{R}^n}\psi(y)dy=1.$$ So $\phi(x)=1$ for all $x$; of course, this $\phi(x)$ is infinitely differentiable. If we also add condition $\phi(x)=0$ for $|x|\geq1$, we break infinite differentiability of $\phi(x)$. This is what seems ambiguous to me.
Nevertheless, in the region, where $\phi(x)=1$, $\mathrm{grad}\phi(x)=0$. Of course, $0\leq\frac{k}{d}$, and this is what you need.