Convergence of Fourier Series — Question 2

Question 2 – Solution

Strategy. A shrinking support can make integral errors small while retaining a fixed peak.

Step 1: Compare the modes of convergence. The peak is always one and direct integration gives $$\boxed{\|h_m\|_\infty=1,\qquad \|h_m\|_2^2=2\int_0^{1/m}(1-mx)^2\,dx=\frac 2{3m}.}$$ The pointwise limit is one at $0$ modulo $2\pi$ and zero elsewhere. Thus $h_m\to 0$ in mean square and almost everywhere, but not at every point and not uniformly. The pointwise limit has a single-point spike in each period.

Step 2: Compute the spectrum. The function is even, so $b_n=0$. Its area is $1/m$, giving mean $1/(2\pi m)$. Integration over its support gives $$\boxed{a_n=\frac 2\pi\int_0^{1/m}(1-mx)\cos nx\,dx =\frac{2m}{\pi n^2}\bigl(1-\cos(n/m)\bigr).}$$ For each fixed $n$, $0\leq a_n\leq 1/(\pi m)$ by $1-\cos u\leq u^2/2$. Therefore every fixed coefficient tends to zero as $m\to\infty$.

Step 3: Keep the target fixed when taking a Fourier limit. For fixed $m$, $|a_n|\leq 4m/(\pi n^2)$, so the series converges absolutely and uniformly. The piecewise smooth Fourier theorem identifies its sum with the continuous target $h_m$. In particular, $$\boxed{\|h_m-S_N^{(m)}\|_\infty\leq\frac{4m}{\pi N}.}$$ The factor $m$ prevents using this bound uniformly over all targets.

Step 4: Compare the iterated limits. For fixed $m$, the inner $N$-limit at zero is $h_m(0)=1$. For fixed $N$, the mean and finitely many coefficients all tend to zero. Thus $$\boxed{\lim_{m\to\infty}\lim_{N\to\infty}S_N^{(m)}(0)=1,\qquad \lim_{N\to\infty}\lim_{m\to\infty}S_N^{(m)}(0)=0.}$$ Fixed-mode convergence does not control the collective contribution of modes whose number grows with $m$. There is no common uniform estimate permitting this interchange.