Fourier Transform 22 | Riemann–Lebesgue Lemma for Fourier Series
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Riemann–Lebesgue lemma states that for every f ∈ L1, the Fourier coefficients f̂(k) satisfy f̂(k) → 0 as k → ±∞.
Briefing
Riemann–Lebesgue lemma pins down a key asymptotic behavior of Fourier series: for any function f in L1, its Fourier coefficients form a sequence that always tends to 0 as the frequency k goes to ±∞. That matters because it guarantees that high-frequency oscillations “wash out” in the Fourier coefficient sense, even when f is not square-integrable—so the result reaches beyond the comfortable L2 setting.
The starting point is the Fourier coefficient formula for an L1 function: the coefficients are defined by an integral and can be packaged as a sequence {f̂(k)} of complex numbers. The lemma asserts that this sequence converges and that its limit is exactly zero. In practical terms, the absolute values |f̂(k)| eventually become arbitrarily small, though the lemma does not quantify how fast the decay occurs.
The proof strategy hinges on a gap: for L2 functions, the conclusion is already known. Parseval’s identity implies that the Fourier coefficients are square-summable, meaning {f̂(k)} lies in ℓ2. Since any square-summable sequence must go to zero at infinity, the lemma follows automatically for f ∈ L2. The real work begins when f is in L1 but not in L2, where f may be unbounded and the square-integrability argument breaks.
To bridge that gap, the argument constructs truncated approximations of f that become square-integrable. For each natural number n, define a 2π-periodic function f_n by “cutting off” the magnitude of f: when |f(x)| ≤ n, keep f(x); when |f(x)| > n, replace it by n (in magnitude). This truncation produces bounded functions, and bounded periodic functions on a finite interval are in L2. As a result, Parseval’s identity applies to each f_n, giving control over the tail of their Fourier coefficients.
Next comes the convergence back to the original f. The proof uses Lebesgue’s dominated convergence theorem: f_n(x) → f(x) almost everywhere as n → ∞, and |f_n(x)| is dominated by |f(x)|, which is integrable because f ∈ L1. Dominated convergence then yields convergence in L1, i.e., ∫|f_n − f| → 0.
That L1 convergence transfers to the Fourier coefficients. For each fixed k, the difference f̂_n(k) − f̂(k) is an integral of (f_n − f) against the complex exponential e^{-ikx}, whose modulus is 1. Taking absolute values and using the triangle inequality shows that |f̂_n(k) − f̂(k)| can be made small uniformly in k once n is large enough.
Finally, the proof combines two tail controls: (1) for large n, the Fourier coefficients of f_n approximate those of f, and (2) for each such f_n, Parseval’s identity forces f̂_n(k) to be small when |k| is large. A careful ε/2 bookkeeping then shows that |f̂(k)| itself becomes smaller than any prescribed ε for all sufficiently large |k|. The lemma is established for complex Fourier coefficients, and the same conclusion follows for real sine/cosine coefficients by standard relationships between the two Fourier representations.
Cornell Notes
Riemann–Lebesgue lemma guarantees that if f is integrable (f ∈ L1), then its Fourier coefficients f̂(k) approach 0 as k → ±∞. The result is immediate for f ∈ L2 because Parseval’s identity makes {f̂(k)} square-summable, and every ℓ2 sequence must vanish at infinity. For f in L1 but not L2, the proof truncates f to bounded functions f_n by capping |f| at n, which puts each f_n in L2. Lebesgue’s dominated convergence theorem shows f_n → f in L1, and that L1 convergence transfers to Fourier coefficients. Combining “approximate by f_n” with “high frequencies of f_n are small” yields f̂(k) → 0.
Why does Parseval’s identity make the lemma automatic for L2 functions?
How does truncating f into f_n help when f is only in L1?
What role does dominated convergence play in the proof?
Why does L1 convergence of f_n − f control Fourier coefficient differences?
How are the two “smallness” steps combined to force f̂(k) → 0?
Review Questions
- What fails when f is in L1 but not in L2, and how does the truncation f_n fix it?
- Where exactly does the modulus |e^{-ikx}| = 1 enter the Fourier coefficient estimate?
- How does the ε/2 argument ensure convergence for both positive and negative k (k → ±∞)?
Key Points
- 1
Riemann–Lebesgue lemma states that for every f ∈ L1, the Fourier coefficients f̂(k) satisfy f̂(k) → 0 as k → ±∞.
- 2
For f ∈ L2, Parseval’s identity implies {f̂(k)} ∈ ℓ2, and ℓ2 sequences necessarily vanish at infinity.
- 3
When f ∈ L1 but not L2, truncating f to bounded functions f_n makes each f_n belong to L2 so Parseval’s identity becomes available.
- 4
Lebesgue’s dominated convergence theorem shows f_n → f in L1 because |f_n| ≤ |f| and f ∈ L1.
- 5
L1 convergence transfers to Fourier coefficients: for fixed k, |f̂_n(k) − f̂(k)| is controlled by ∫|f_n − f|.
- 6
A two-step ε/2 argument combines approximation by f_n with the high-frequency decay of f̂_n(k) to conclude f̂(k) → 0.
- 7
The lemma holds for complex Fourier coefficients and therefore also for real sine/cosine coefficients via standard Fourier-series relationships.