- #1

- 210

- 10

## Main Question or Discussion Point

Where is the fallacy in this "proof" that the Fourier series of f(x) converges to f(x) if f is continuous at x and has period 2π? (I read in Wikipedia that a counterexample had been provided).

Start with the Dirichlet integral for the N-th partial sum of the (trigonometric) Fourier series. After some trig manipulations, u-substitutions and applications of the Riemann-Lebesgue lemma on intervals where the integrand is bounded, it is possible to write:

[tex] \lim_{N \rightarrow ∞}[s_N(x)-f(x)] = \lim_{N \rightarrow ∞} \frac{1}{\pi}\int_{-\delta}^{\delta}[f(x+\alpha)-f(x)]\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) d\alpha. [/tex]

for any δ > 0.

(this part is proven in standard textbooks and the fallacy most likely isn't here, and I am convinced that all steps are true. For a reference, I read it in N. Piskounov's "Calcul Differentiel et Integral", Tome II, Ch. XVII, #9. in french).

Now whereas the book assumes more restrictive conditions on f(x) to prove convergence of the series, I was wondering whether mere continuity was sufficient. Here's my attempt at a proof, which is supposedly false:

Given any ε > 0, choose N = 1/ε. Then choose a δ > 0 such that |f(x+α)-f(x)| < 1/N^2 = ε^2 for all |α| < δ (we are allowed to do so if f is continuous at x, since N is a fixed number now). Use this δ in the integral formula derived above. Moreover, [itex] 0 < \frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) ≤ N[/itex] on some interval around 0 (assume δ is included in it).

Then [itex] |s_N(x) - f(x)| ≤ \frac{1}{\pi}\int_{-\delta}^{\delta}|f(x+\alpha)-f(x)|\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) d\alpha ≤ \frac{1}{\pi}\int_{-\delta}^{\delta}\frac{1}{N^2}N d\alpha = \frac{2\delta}{\pi N} < ε [/itex] if δ is also smaller than 1/2.

We showed that the absolute value of the difference can be made as small as desired by requiring N to be large enough. This proves that the limit as N goes to infinity exists and that s_N(x) converges to f(x) for all x at which f is continuous.

Start with the Dirichlet integral for the N-th partial sum of the (trigonometric) Fourier series. After some trig manipulations, u-substitutions and applications of the Riemann-Lebesgue lemma on intervals where the integrand is bounded, it is possible to write:

[tex] \lim_{N \rightarrow ∞}[s_N(x)-f(x)] = \lim_{N \rightarrow ∞} \frac{1}{\pi}\int_{-\delta}^{\delta}[f(x+\alpha)-f(x)]\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) d\alpha. [/tex]

for any δ > 0.

(this part is proven in standard textbooks and the fallacy most likely isn't here, and I am convinced that all steps are true. For a reference, I read it in N. Piskounov's "Calcul Differentiel et Integral", Tome II, Ch. XVII, #9. in french).

Now whereas the book assumes more restrictive conditions on f(x) to prove convergence of the series, I was wondering whether mere continuity was sufficient. Here's my attempt at a proof, which is supposedly false:

Given any ε > 0, choose N = 1/ε. Then choose a δ > 0 such that |f(x+α)-f(x)| < 1/N^2 = ε^2 for all |α| < δ (we are allowed to do so if f is continuous at x, since N is a fixed number now). Use this δ in the integral formula derived above. Moreover, [itex] 0 < \frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) ≤ N[/itex] on some interval around 0 (assume δ is included in it).

Then [itex] |s_N(x) - f(x)| ≤ \frac{1}{\pi}\int_{-\delta}^{\delta}|f(x+\alpha)-f(x)|\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}\sin (N\alpha) d\alpha ≤ \frac{1}{\pi}\int_{-\delta}^{\delta}\frac{1}{N^2}N d\alpha = \frac{2\delta}{\pi N} < ε [/itex] if δ is also smaller than 1/2.

We showed that the absolute value of the difference can be made as small as desired by requiring N to be large enough. This proves that the limit as N goes to infinity exists and that s_N(x) converges to f(x) for all x at which f is continuous.