Need help on clarifying a step in proof of Plancherel's theorem in Evans' book PDE

I was stuck when I read the proof of Plancherel's theorem in Line 9, Page 188 of Evans' book Partial Differential Equations, 2nd Edition. Evans wrote there (I quote here):

$~~~~\text{2. }$ Now take $u\in L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$ and set $v(x):=\bar{u}(-x).$ Let $w:=u*v\in L^1(\mathbb{R}^n)\cap C(\mathbb{R}^n)$ and check ...

I do not understand why $u*v$ is in $C(\mathbb{R}^n).$ I had tried this: Let $x, x_1\in\mathbb{R}^n,$ then \begin{align*} |w(x)-w(x_1)|&=\left|\int_{\mathbb{R}^n} \Big(u(x-y)v(y)-u(x_1-y)v(y)\Big)dy\right|\\ &\leq\int_{\mathbb{R}^n}\Big|u(x-y)-u(x_1-y)\Big||v(y)|dy, \end{align*} but how could I derive from this that \begin{align*} \int_{\mathbb{R}^n}\Big|u(x-y)-u(x_1-y)\Big||v(y)|dy\to 0, \qquad\text{ as } x_1\to x? \end{align*}

Can anyone help me clarify this, or where can I find some references about this kind of continuity property of convolution?

• You have said that $u\in L^2({\mathbb R}^n)\cap L^2({\mathbb R}^n)$, should that be two different properties? Also, don't forget that $v(x)$ is a specific function of $u(x)$ May 4, 2014 at 6:21
• Sorry, I have made a typing mistake. May 7, 2014 at 5:04

I think your initial assumption is $u \in L^2(\mathbb{R}^n) \cap L^1(\mathbb{R}^n)$. Anyway the crucial part here is that translations are continuous in the $L^p$ norm if $1 \leq p < +\infty$, which makes your estimate easy. A more general form of what you need here is that if $f \in L^p$ and $g \in L^q$ with $p$ and $q$ conjugate exponents, $f*g$ is bounded and uniformly continuous. Real analysis by Folland (along with many other books) has all of these facts in detail.