Let \(\varphi : \mathbb{R} \to \mathbb{R}\) be a non-negative, \(C^\infty\) function with compact support such that
\[\int_{\mathbb{R}} \varphi(x) \, dx = 1.\]
Define
\[\varphi_\sigma(x) = \sigma^{-1} \varphi\left(\frac{x}{\sigma}\right) \quad \text{and} \quad u_\sigma(x) = \int \varphi_\sigma(x - y) u(y) \, dy.\]
For \(u \in L^2(\mathbb{R})\), prove
\[\int_{\mathbb{R}} |u_\sigma(x)|^2 \, dx \leq \int_{\mathbb{R}} |u(x)|^2 \, dx.\]
Solution
1. Regard the mollifier as a probability density.
Since \(\varphi\ge0\) and \[ \int_{\mathbb R}\varphi=1, \] the rescaled function \(\varphi_\sigma\) is nonnegative and satisfies \[ \int_{\mathbb R}\varphi_\sigma(y)\,dy=1. \] Thus, for each fixed \(x\), the measure \[ d\mu_x(y):=\varphi_\sigma(x-y)\,dy \] is a probability measure on \(\mathbb R\).
2. Apply Jensen’s inequality pointwise.
By Jensen’s inequality for the convex function \(z\mapsto |z|^2\), \[ \begin{aligned} |u_\sigma(x)|^2 &=\left|\int_{\mathbb R}u(y)\varphi_\sigma(x-y)\,dy\right|^2\\ &\le \int_{\mathbb R}|u(y)|^2\varphi_\sigma(x-y)\,dy. \end{aligned} \]
3. Integrate in \(x\).
Tonelli’s theorem gives \[ \begin{aligned} \int_{\mathbb R}|u_\sigma(x)|^2\,dx &\le \int_{\mathbb R}\int_{\mathbb R}|u(y)|^2 \varphi_\sigma(x-y)\,dy\,dx\\ &=\int_{\mathbb R}|u(y)|^2 \left(\int_{\mathbb R}\varphi_\sigma(x-y)\,dx\right)dy\\ &=\int_{\mathbb R}|u(y)|^2\,dy. \end{aligned} \] Therefore \[ \boxed{\|u_\sigma\|_2\le\|u\|_2.} \]