求证：对所有$f \in L^{1}(\mathbb R)$，$||f*\mu||_1=||f||_1$时$\mu$退化

Alright, let's break down this problem and work through a rigorous proof step by step. We need to show that if a complex Borel measure $\mu$ on $\mathbb{R}$ satisfies $|f * \mu|_1 = |f|_1$ for every $f \in L^1(\mathbb{R})$, then $\mu$ must be a degenerate measure (i.e., $\mu = c\delta_a$ where $|c|=1$ and $\delta_a$ is the Dirac delta measure at some $a \in \mathbb{R}$).

Step 1: Prove $|\mu|_1 = 1$ #

First, recall Young's inequality: for any $f \in L^1(\mathbb{R})$ and complex Borel measure $\mu$, we have $|f * \mu|_1 \leq |f|_1 |\mu|_1$. Our hypothesis tells us $|f * \mu|_1 = |f|_1$, so dividing both sides by $|f|_1$ (for $f \neq 0$) gives $|\mu|_1 \geq 1$.

Now take a non-negative approximate identity ${f_n}$: these are $L^1$ functions with $|f_n|_1 = 1$, $f_n \geq 0$, and $\int f_n(x)g(x)dx \to g(0)$ for all continuous bounded $g$. By hypothesis, $|f_n * \mu|1 = 1$. But we also have:
$$
|f_n * \mu|1 = \int \left| \int f_n(x-y)d\mu(y) \right| dx \leq \int \int f_n(x-y)dx |d\mu(y)| = \int |d\mu(y)| = |\mu|1
$$
To show $|\mu|1 = 1$, suppose for contradiction that $|\mu|1 > 1$. Then there exists a continuous bounded function $g$ with $|g|\infty = 1$ such that $\left| \int g d\mu \right| > 1$. But $f_n * g$ converges uniformly to $g$ (since $f_n$ is an approximate identity), so:
$$
\left| \int g d\mu \right| = \lim{n \to \infty} \left| \int (f_n * g)(y)d\mu(y) \right| = \lim{n \to \infty} \left| \int g(x)(f_n * \mu)(x)dx \right| \leq \limsup{n \to \infty} |g|\infty |f_n * \mu|_1 = 1
$$
This contradiction forces $|\mu|_1 = 1$.

Step 2: Show $|\widehat{\mu}(\xi)| = 1$ for all $\xi \in \mathbb{R}$ #

Recall that the Fourier transform of $f * \mu$ is $\widehat{f}(\xi)\widehat{\mu}(\xi)$, where $\widehat{\mu}(\xi) = \int e^{-i\xi y}d\mu(y)$ (a continuous function, since $\mu$ is a regular Borel measure).

Take any $\xi_0 \in \mathbb{R}$, and define $f(x) = e^{i\xi_0 x}h(x)$ where $h \in L^1(\mathbb{R})$ and $|h|_1 = 1$. Then $|f|_1 = 1$, so $|f * \mu|1 = 1$. By the Fourier transform property, $|\widehat{f * \mu}|\infty \leq |f * \mu|_1 = 1$.

Now take $h$ to be an approximate identity ${h_n}$: $\widehat{h_n}(\eta) \to 1$ for all $\eta$ as $n \to \infty$. Then $\widehat{f * \mu}(\xi) = \widehat{h_n}(\xi - \xi_0)\widehat{\mu}(\xi) \to \widehat{\mu}(\xi)$ pointwise. Since $|\widehat{f * \mu}|_\infty \leq 1$, we get $|\widehat{\mu}(\xi_0)| \leq 1$.

To show equality, suppose for contradiction that there exists $\xi_0$ with $|\widehat{\mu}(\xi_0)| < 1$. By continuity of $\widehat{\mu}$, there's an interval $I = (\xi_0 - \delta, \xi_0 + \delta)$ where $|\widehat{\mu}(\xi)| \leq r < 1$ for all $\xi \in I$. Construct an $L^1$ function $h_n$ whose Fourier transform $\widehat{h_n}$ is $1$ on $I$, $0$ outside $I + [-\frac{1}{n}, \frac{1}{n}]$, and smooth in between. Normalize $h_n$ so $|h_n|_1 = 1$.

Let $f_n(x) = e^{i\xi_0 x}h_n(x)$: $|f_n|_1 = 1$, so $|f_n * \mu|_1 = 1$. But:
$$
\widehat{f_n * \mu}(\xi) = \widehat{h_n}(\xi - \xi_0)\widehat{\mu}(\xi)
$$
For $\xi \in \xi_0 + I$, $\widehat{h_n}(\xi - \xi_0) = 1$, so $|\widehat{f_n * \mu}(\xi)| = |\widehat{\mu}(\xi)| \leq r < 1$. For $\xi \notin \xi_0 + I$, $\widehat{h_n}(\xi - \xi_0) = 0$ or bounded by $1$, so $|\widehat{f_n * \mu}(\xi)| \leq 1$.

Using the inverse Fourier transform bound:
$$
|f_n * \mu|_1 \leq \frac{1}{2\pi} \int |\widehat{f_n * \mu}(\xi)|d\xi \leq \frac{1}{2\pi}\left( r|I| + (2\pi - |I|) \right) = 1 - \frac{|I|(1 - r)}{2\pi} < 1
$$
This contradicts $|f_n * \mu|_1 = 1$, so $|\widehat{\mu}(\xi)| = 1$ for all $\xi$.

Step 3: Conclude $\mu$ is degenerate #

We know $\widehat{\mu}$ is continuous, $|\widehat{\mu}(\xi)| = 1$ for all $\xi$, and $|\mu|1 = 1$. Let $c = \widehat{\mu}(0) = \int d\mu$ (so $|c| = 1$), and define $\nu = \overline{c}\tau{a}\mu$ where $a$ is chosen such that $\widehat{\mu}(\xi) = c e^{-i a \xi}$ (since $\widehat{\mu}$ is a continuous function of modulus 1, it must be a character of $\mathbb{R}$, i.e., $e^{-i a \xi}$ scaled by $c$).

Then $\widehat{\nu}(\xi) = 1$ for all $\xi$, meaning $\int e^{-i\xi y}d\nu(y) = 1$ for all $\xi$. This implies $\int (e^{-i\xi y} - 1)d\nu(y) = 0$ for all $\xi$. By the Stone-Weierstrass theorem, trigonometric polynomials are dense in the space of continuous bounded functions on $\mathbb{R}$, so $\int g d\nu = g(0)$ for all continuous bounded $g$. Thus $\nu = \delta_0$, so $\mu = c\delta_a$, a degenerate measure.

Note on Translation Continuity #

As you mentioned, David Ullrich's approach used translation continuity in $L^1$, but measure translation is continuous in weak convergence but not in total variation. Our proof avoids this pitfall by leveraging Fourier transform properties and approximate identities, which let us work with pointwise convergence of Fourier transforms and avoid directly handling total variation discontinuities.

内容的提问来源于stack exchange，提问作者Kavi Rama Murthy