Convergence of Fourier Series — Question 1

Question 1 – Solution

Strategy. Distinguish the assigned function, its Fourier pointwise limit and the equivalence class measured by an integral norm.

Step 1: Compute coefficients and limits. Apart from isolated values, $f=1/2+(3/2)\operatorname{sgn}(\sin x)$. Thus $a_0=1$, every positive cosine coefficient is zero, and $$\boxed{b_n=\frac{3(1-(-1)^n)}{\pi n} =\begin{cases}6/(\pi n),&n\text{ odd},\\0,&n\text{ even}.\end{cases}}$$ The series sum equals $-1$ or $2$ on the open intervals and $1/2$ at every jump. It does not equal either assigned value $7$ or $-4$ there.

Step 2: Verify mean-square convergence. Isolated values do not affect integrals. Parseval and orthogonality give $$\begin{gathered} \boxed{\|f-S_N\|_2^2=\frac{36}{\pi} \sum_{\substack{n>N\\n\ {\mathrm{odd}}}}\frac 1{n^2} \leq\frac{36}{\pi N}\longrightarrow 0.} \end{gathered}$$ The coefficient tail is square summable despite failure at the assigned jump values.

Step 3: Separate point and essential errors. Every $S_N(0)=1/2$, so $\|f-S_N\|_\infty\geq 13/2$ for every $N$. Moreover, by continuity of $S_N$, as $x\downarrow 0$ its error against the right-hand value $2$ tends to $3/2$. For any $\varepsilon>0$, that error exceeds $3/2-\varepsilon$ on an interval of positive length. Hence $$\boxed{\operatorname*{ess\,sup}|f-S_N|\geq 3/2.}$$ Removing isolated exceptional values cannot remove this obstruction.

Step 4: Test possible redefinitions. The one-sided limits remain $-1$ and $2$ after any change only at the jumps. The resulting target is still discontinuous. A uniform limit of the continuous trigonometric polynomials $S_N$ would be continuous, so no such redefinition restores uniform convergence. Mean-square convergence and the open-interval pointwise limits are unchanged. Filled points in the graph show the assigned values.