求证：若f∈C(𝕋)且∑ₙ|f̂(n)||n|ˡ<∞，则f∈Cˡ(𝕋)

Hey there! Let's walk through this problem clearly, and we'll make sense of that "differentiation under the integral sign" hint along the way.

First, Recap Key Background #

• Since $f \in C(\mathbb{T})$, its Fourier series is $f(x) = \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{inx}$.
• The given condition $\sum_{n \in \mathbb{Z}} |\hat{f}(n)| |n|^l < \infty$ is a strong convergence condition—way stronger than just the Fourier series converging pointwise to $f$.

Step 1: Start with $l=1$ (First-Order Differentiability) #

Let's build intuition with the simplest case, then generalize to $l$ orders.

1. Define the逐项求导 partial sums:
   Let $S_k(x) = \sum_{|n| \leq k} \hat{f}(n) e^{inx}$ be the $k$-th partial sum of $f$'s Fourier series. If we formally differentiate term-by-term, we get:
   $$g_k(x) = S_k'(x) = \sum_{|n| \leq k} \hat{f}(n) (in) e^{inx}$$

2. Prove $g_k$ converges uniformly to a continuous function:
   By the Weierstrass M-Test, we just need to bound the absolute value of each term and show the sum of bounds converges. The absolute value of each term in $g_k$ is:
   $$|\hat{f}(n) (in) e^{inx}| = |\hat{f}(n)| |n|$$
   The given condition tells us $\sum_{n \in \mathbb{Z}} |\hat{f}(n)| |n| < \infty$ (since $l=1$ here). So $g_k$ converges uniformly to some $g \in C(\mathbb{T})$ (uniform limits of continuous functions are continuous).

3. Use integration to link $g$ to $f'$ (the integral hint!):
   Now, integrate $g_k$ from $0$ to $x$:
   $$\int_0^x g_k(t) dt = \int_0^x S_k'(t) dt = S_k(x) - S_k(0)$$
   Since $S_k$ converges uniformly to $f$ (because $\sum |\hat{f}(n)| \leq |\hat{f}(0)| + \sum_{n \neq 0} |\hat{f}(n)| |n| < \infty$, so again Weierstrass M-Test applies), we can swap the limit and integral:
   $$\int_0^x g(t) dt = \lim_{k \to \infty} \int_0^x g_k(t) dt = \lim_{k \to \infty} [S_k(x) - S_k(0)] = f(x) - f(0)$$
   Now, apply the Fundamental Theorem of Calculus: since $g$ is continuous, differentiating both sides gives $g(x) = f'(x)$. And since $g$ is continuous, $f' \in C(\mathbb{T})$—so $f \in C^1(\mathbb{T})$.

Step 2: Generalize to $l$ Orders via Induction #

Now we can extend this to any positive integer $l$ using induction:

• Base case: We just proved $l=1$ works.
• Inductive step: Suppose for some $m \geq 1$, if $\sum_{n} |\hat{f}(n)| |n|^m < \infty$, then $f \in C^m(\mathbb{T})$, and the $m$-th derivative $f^{(m)}$ has Fourier coefficients $\hat{f}(n) (in)^m$.
• For $l=m+1$: The given condition is $\sum_{n} |\hat{f}(n)| |n|^{m+1} = \sum_{n} |\hat{f}(n) (in)^m| |n| < \infty$. By the inductive hypothesis, $f^{(m)} \in C(\mathbb{T})$, and its Fourier coefficients are $\hat{f}(n) (in)^m$. Applying the $l=1$ case to $f^{(m)}$, we find $f^{(m)} \in C^1(\mathbb{T})$—which means $f \in C^{m+1}(\mathbb{T})$.

Why the "Differentiation Under the Integral Sign" Hint Matters #

The hint is pointing to the key step where we swap the limit (as $k \to \infty$) with the integral of $g_k(t)$. This is essentially the reverse of differentiation under the integral sign: since we know the partial sums' derivatives converge uniformly, we can justify integrating the limit instead of integrating each partial sum first. This link between the uniformly convergent derivative series and the original function's derivative is what bridges the Fourier series condition to $l$-th order continuous differentiability.

内容的提问来源于stack exchange，提问作者paarth