Convergence of Fourier Series — Question 10

Question 10 – Solution

Strategy. Damping controls all high frequencies for positive $t$; removing it requires a separate convergence argument.

Step 1: Prove smoothness for positive damping. Parseval gives $E=\sum(a_n^2+b_n^2)<\infty$. After $k$ derivatives, the absolute value of the $n$th harmonic is at most $n^ke^{-n^2t}\sqrt{a_n^2+b_n^2}$. Cauchy–Schwarz bounds the sum by $$\sqrt E\left(\sum_{n\geq 1}n^{2k}e^{-2n^2t}\right)^{1/2}<\infty.$$ The exponential dominates every power, so the Weierstrass test applies for each fixed $t>0$ and each integer $k\geq 0$. Repeated use of the uniform derivative theorem justifies differentiation; $U_t$ is smooth and periodic.

Step 2: Remove damping in the integral norm. The Fourier coefficients of the difference give $$\boxed{\|U_t-f\|_2^2=\pi\sum_{n\geq 1} (1-e^{-n^2t})^2(a_n^2+b_n^2).}$$ Given $\varepsilon>0$, choose $K$ so that $\pi\sum_{n>K}(a_n^2+b_n^2) <\varepsilon/2$. Since $0\leq 1-e^{-n^2t}\leq 1$, this bounds the error tail for every $t$. The remaining finite sum tends to zero, hence is below $\varepsilon/2$ for sufficiently small $t$. This proves the limit.

Step 3: Obtain a quantitative rate under extra information. For $z\geq 0$, $1-e^{-z}\leq\min(z,1)$, so $(1-e^{-z})^2\leq z$. Applying this with $z=n^2t$ gives $$\boxed{\|U_t-f\|_2^2\leq\pi t\sum n^2(a_n^2+b_n^2)\leq\pi Mt.}$$ Taking square roots yields the requested rate; this uses a stronger hypothesis than mere square summability.

Step 4: Separate smoothing from uniform recovery. Uniform integration of the damped series preserves the mean $m$. Parseval gives $\|U_t-m\|_2^2=\pi\sum e^{-2n^2t}(a_n^2+b_n^2)\leq\|f-m\|_2^2$. Every $U_t$ is continuous. Choose an $L^2$ periodic step function with a jump: it cannot be a uniform limit of these continuous functions. Even isolated redefinitions of $f$ are invisible to the coefficients. Thus smooth approximants and mean-square recovery do not alone guarantee uniform recovery of an arbitrarily assigned representative.