I have been working on the following exercise from Stein's functional analysis textbook. It states
Show that a bounded function $f$ on $\mathbb{R}^d$ satisfies the following Lipschitz condition $$|f(x)- f(y)| \leq C|x-y|, \;\;\;\forall x,y \in \mathbb{R}^d, \;\; C>0$$ if and only if $f \in L^\infty (\mathbb{R}^d)$ and all the first order partial derivative $\frac{\partial f}{\partial x^j}$ (defined in the sense of distributions if necessary), $1 \leq j \leq d$, belong to $L^\infty (\mathbb{R}^d)$.
Hint: To prove this you may let $f_n = f*\phi_n$, where $\phi_n$ is an approximation to the identity, e.g. $\phi \in \mathcal{D}(\mathbb{R}^d)$ such that $\int_{\mathbb{R}^d}\phi = 1$ and set $\phi_n(x) = n^d \phi(nx)$. Then you can prove that $\frac{\partial f_n}{\partial x^j} \in L^\infty (\mathbb{R}^d)$ uniformly in $n$.
Note that $\mathcal{D}(\mathbb{R}^d)$ is the space of complex valued infinitely differentiable functions with compact support on $\mathbb{R}^d$.
I have been able to prove the forward direction. However, how am I supposed to use the given hint to prove the reverse direction? To prove the hint, my thoughts are:
Consider the distribution given by
$$\int_{\mathbb{R}^d}f(x) \phi(x) dx$$
where $\phi \in \mathcal{D}(\mathbb{R}^d)$. Note then that
$$f_n(x) = (f*\phi_n)(x) = \int_{\mathbb{R}^d} f(x-y) \phi_n(y) dy = \int_{\mathbb{R}^d} f(x-y) n^d\phi(ny) dy = n^d \int_{\mathbb{R}^d} f(x-y)\phi(ny) dy$$
$$\Rightarrow \frac{\partial f_n}{\partial x^j}(x) = n^d\int_{\mathbb{R}^d}\frac{\partial f}{\partial x^j}(x-y) \phi(ny) dy, \;\; \text{(derivative of distribution)}$$
$$\Rightarrow \Big|\frac{\partial f_n}{\partial x^j}(x) \Big| = n^d \Big| \int_{\mathbb{R}^d} \frac{\partial f}{\partial x^j}(x-y) \phi(ny)dy\Big|$$
Since we assumed that $\frac{\partial f}{\partial x^j} \in L^\infty(\mathbb{R}^d)$, and $\phi$ is an approximation to the identity, it should follow that $\frac{\partial f_n}{\partial x^j} \in L^\infty$. Now, how do I use this hint to prove the Lipschitz condition?