If $f$ is a continuous function and $\phi$ is a smooth function with compact support such that $\int\phi\,dx=1$, is $f*\phi$ Lipschitz continuous?

Question

Let us take a continuous function $f$. Let $\phi$ be a smooth function with a compact support, such that $\int \phi(x)\,dx=1$. We consider the following convolution

$(f*\phi)(x)=\int f(x-y)\phi(y)\,dy$. Is such a function Lipschitz continuous?

I know this function is smooth, okay. But how do I know its first derivative is bounded? I have only information about $\phi$ itself and I know it has a compact support, hence it is bounded. I suppose that it means that its derivative also is bounded in this way but can you help me? Or maybe it can be shown strictly from definition?

Answer 1

The derivative of $\phi$ is also compactly supported and continuous by the smoothness of $\phi$, hence it is bounded. However, this does not mean that $f *\phi$ is Lipschitz, but, as the first sentence suggests, this will depend on $f$, not on $\phi$.

If $f(x)=x^2$, $\phi$ compactly supported on $[0,1]$ and positive, thus, for $x \ge1$ :

$$f * \phi(x) = \int_{0}^{1} (x-y)^2 \phi(y) dy \ge \int_{0}^{1} (x-1)^2 \phi(y) dy.$$

Thus, since $\int_0^1 \phi(y) dy = 1$ :

$$f * \phi(x) \ge (x-1)^2$$

and $f*\phi$ cannot be Lipschitz continuous on $\mathbb{R}$.

However, $f * \phi$ is locally Lipschitz as $(f*\phi)'=f* \phi'$ and this derivative is bounded on any compact set by continuity of $f$ and boundedness of $\phi'$.

Answer 2

You can use the following facts: $$ D^\alpha(f \ast \phi) = f \ast D^\alpha\phi, \qquad \|f \ast D^\alpha\phi\|_\infty \leq \|f\|_p \|D^\alpha\phi\|_{p'}\,. $$ Hence, if $f\in L^p$ for some $p$, you can conclude that $f\ast\phi$ has bounded derivatives.