5.5 Convergence of Fourier Series
Everything so far has been formal. The coefficients were computed as though the series converged and as though term-by-term integration were legitimate, and neither was justified. The Riemann–Lebesgue lemma and the Dirichlet kernel of the previous section are the tools that settle it, and the theorem below is what they were assembled for.
Theorem 5.5.1. Let \(f(x)\) be Riemann integrable on \([-\pi ,\pi ]\) and let it be extended periodically outside this interval. Suppose that at a point \(x\), \(f(x)\) satisfies the following conditions:
- (i).
- Both \(f(x^-)\) and \(f(x^+)\) exist, where \(f(x^-)\) and \(f(x^+)\) are left sided and right-sided limits of \(f(x)\) at \(x\), and \[f(x^+)= \frac {1}{2}\left [f(x^-) + f(x^+)\right ];\]
- (ii).
- Both one-sided derivatives \[f'(x^-) = \lim _{h\rightarrow 0^-} \frac {f(x + h) - f(x^-)}{h},\] exist. Then the Fourier series of \(f(x)\) converges to \(f(x)\) at \(x\), that is \[\lim _{n\rightarrow \infty } S_n(x) = \begin {cases} f(x) , & \text {if}\,\, x\,\, \text {is a of continuity}\\ \frac {1}{2}\left [f(x^+) + f(x^-)\right ], & \text {if}\,\, x\,\, \text {is a continuity} \end {cases}\]
Proof. We have \begin {align*} S_n(x) & = \frac {a_0}{2} + \sum ^n_{k = 1} \left [a_k\cos kx + b_k\sin kx\right ]\\ & = \frac {1}{2\pi }\int _{-\pi }^{\pi } f(t)\, dt + \sum ^n_{k = 1} \left [\frac {1}{\pi }\left (\int _{-\pi }^{\pi }f(t)\, \cos kt\, dt\right ) \cos kx + \frac {1}{\pi }\left (\int _{-\pi }^{\pi }f(t)\, \sin t\, dt\right ) \sin kx\right ]\\ & = \frac {1}{\pi }\int _{-\pi }^{\pi }f(t)\left [\frac {1}{2} + \sum _{k = 1}^n\left (\cos kt \, \cos kx + \sin kt\, \sin kx\right )\right ] \, dt\\ & = \frac {-1}{\pi }\int _{-\pi }^{\pi } f(t)\left [\frac {1}{2} + \sum _{k = 1}^n\cos (t - x)\right ] \, dt\\ & = \frac {1}{\pi }\int _{-\pi }^{\pi } f(t) \left [\frac {\sin \left [\left (\frac {n+ 1}{2}\right )(t - x)\right ]}{2\sin \left (\frac {t - x}{2}\right )}\right ]\, dt \end {align*}
\(2\pi \), so that their integral will be the same over \(-\pi -x\) to \(\pi -x\) and over \(-\pi \) to \(\pi \). Hence \begin {equation} S_n(x) = \int _{-\pi }^{\pi } f(x + u)\, \frac {\sin \left [\left (n + \frac {1}{2}\right )u\right ]}{2\sin \frac {u}{2}} \end {equation} we must show that \[\lim _{n\rightarrow \infty }S_n(x)\] □
Definition 5.5.2. A function \(f(x)\) is said to be piecewise continuous on \([a,b]\) if it is continuous on \([a,b]\) except for a finite number of discontinuities of the first kind in \([a,b]\), and, in addition both \(f(a^+)\) and \(f(b^-)\) exist.
Corollary 5.5.3. Suppose that \(f(x)\) is piecewise continuous on \([-\pi ,\pi ]\), and that it can be extended periodically outside this interval, in addition, if, at each interior point of \([-\pi ,\pi ]\), \(f'(x^+)\) and \(f'(x^-)\) exist and \(f'(-\pi ^+)\) and \(f'(-\pi ^-)\) exist, then at a point \(x\), the Fourier series of \(f(x)\) converges to \(\frac {1}{2}[f(x^-) + f(x^+)]\).
Proof. Follows from Theorem 5.5.1 since a piecewise continuous function is Riemann integrable. □
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