If $\phi$ vanishes outside of $|x| < 2$, $\phi = 1$ where $|x| < 1$, prove/disprove $f * \phi$ is in $L^1$/$L^2$

Question

Problem Statement: Let $\phi$ be a positive, smooth function. Suppose $\phi$ vanishes outside a compact subset of $\{x : |x| < 2\}$ and satisfies $\phi(x) = 1$ if $|x| < 1$. Let $f$ be a function in $L^2$. If $f \in L^1$, is $f * \phi \in L^1$? If $f \in L^2$, is $f * \phi \in L^2$?

My attempt at a solution: So, I started by writing out

$$\int \left| \int f(x-y)\phi(y) \ dy\right|dx \le \int \int |f(x-y)||\phi(y)|\ dy \ dx$$ and $$\int \left| \int f(x-y)\phi(y) \ dy\right|^2 dx \le \int \left( \int |f(x-y)||\phi(y)|\ dy\right)^2 \ dx$$

and I've tried messing around with Holder's inequality to bound these norms. However, though I think if I knew that if $\phi$ was in $L^1$ or $L^2$, then I could get a bound, I'm not sure what the restrictions on $\phi$ tell me about the convolution. I guess that if the compact subset that $phi$ vanished on was $[-1,1]$, then I would be able to say that the convolution was in $L^1$ if $f \in L^1$, and $L^2$ if $f \in L^2$. But, the compact subset could be larger than that, and I'm unsure whether the smoothness of $\phi$ gets me a nice bound...

Answer 1

Both results follow immediately with an application of

Young's inequality Let $1 \le p,q,r \le \infty$ be such that $$\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1.$$ If $f \in L^p$ and $g \in L^q$ then $f * g \in L^r$ and we get the following bound on the norm $$\|f * g\|_r \le \|f\|_p\|g\|_q.$$

To answer your first question, let $f \in L^1$ and notice that if $\phi \in C^{\infty}_c$ then $\phi \in L^1$. Indeed $$\int |\phi| = \int_{\text{supp}(\phi)}|\phi| \le \|\phi\|_{\infty}\lambda(\text{supp}(\phi)) < \infty.$$ Then Young's inequality applies yielding $$\|f * \phi\|_1 \le \|f\|_1\|\phi\|_1.$$

To answer your second question, apply Young's inequality with $p = r = 2$, $q = 1$.

---

---

If you don't want to apply such a powerful result we can prove everything by hand (keep in mind that $C^{\infty}_c \subset L^1$).

Indeed, all is needed in the first case is an application of Tonelli's Theorem:

\begin{align} \int \left| \int f(x-y)\phi(y)\,dy\right|\,dx \le & \int \int |f(x-y)||\phi(y)|\color{blue}{\,dy \,dx} = \int \int |f(x-y)||\phi(y)|\color{blue}{\,dx \,dy} \\ = & \int |\phi(y)| \int |f(x - y)|\,dx\,dy = \|f\|_1\int |\phi(y)|\,dy = \|f\|_1\|\phi\|_1. \end{align}

For $f \in L^2$ we can apply Minkowski's integral inequality. Let's see how it goes:

\begin{align} \Big(\int \Big|\int f(x - y)\phi(y)\,dy\Big|^2\,dx\Big)^{\frac{1}{2}} \le & \int \Big(\int |f(x - y)|^2|\phi(y)|^2\,dx\Big)^{\frac{1}{2}}\,dy \\ = & \int |\phi(y)|\Big(\int |f(x - y)|^2\,dx\Big)^{\frac{1}{2}}\,dy = \|f\|_2\|\phi\|_1. \end{align}

Hopefully this answers your questions.