Lecture 16: Fejer’s Theorem and Convergence of Fourier Series

April 15, 2021 We’ll continue the discussion of Fourier series today – last time, we defined the Fourier coefficients Z π 1 fˆ(n) = f (t)e −int dt 2π −π for any function f ∈ L2 ([−π, π]), which we can think of as the L2 inner product of f with e −int up to a constant. Defining the Nth partial sums SN f (x) = N X fˆ(n)e inx , n=−N we wanted to know whether SN f always converges to f in L2 – that is, whether for all f ∈ L2 ([−π, π]) we have limN→∞ ||f − SN f ||2 = 0. 2 Based on our discussion of Hilbert spaces, this question is equivalent to asking whether a function ([−π, π]) n f ∈Lo 1 inx is a with all Fourier coefficients zero must be the zero function (since we’re trying to ask whether √2π e n∈Z maximal orthonormal subset). Our main step last time was to define the Cesaro-Fourier mean N σN f (x) = 1 X Sk f (x), N +1 k=0 hoping that means of sequences converge better than the sequences themselves. Our goal is then to show that ||σN f − f ||2 → 0 as N → ∞, and that will give us the desired convergence result for Fourier series. We’ll first rewrite the partial Fourier sums slightly differently, much like how we previously used the Dirichlet kernel: Proposition 173 For all f ∈ L2 ([−π, π]), we have Z π KN (x − t)f (t)dt, σN f (x) = KN (x) = −π    N+1 2π 1   2π(N+1) x =0  x) sin( N+1 2 sin x2 2 otherwise. The function KN (x) is called the Fejér kernel, and it has the following properties: (1) KN (x) ≥ 0 and KN (x) = R∞ KN (−x) for all x, (2) KN is periodic with period 2π, (3) −∞ KN (t)dt = 1, and (4) for any δ ∈ (0, π) and for all δ ≤ |x| ≤ π, we have |KN (x)| ≤ 1 2π(N+1) sin2 δ 2 . The idea is that the Fejér kernel grows more and more concentrated at the origin as N → ∞, but the area of the curve is always 1 (like the physics Dirac delta function) – here’s a picture for N = 8: 83 The reason we might believe that these Cesaro means converge to f is that Z π σN f (x) = KN (x − t)f (t)dt, −π and KN is very sharply peaked around t = x, so as N gets larger and larger, the main contribution to the integral comes from f (x) ≈ f (t) if f is well-behaved enough. So then we end up with Z π ≈ f (x) KN (x − t)dt = f (x) · 1, −π since kN evaluates to the same over any interval of length 2π by periodicity. So that’s a heuristic motivation for working with the Cesaro means here! (Some of these properties also applied when we did a similar procedure with our partial sums SN f (x), but the Dirichlet kernel is not nonnegative – that difference actually makes a big difference in the final proof.) Proof. Recall that Z π Dk (x − t)f (t)dt Sk f (x) = −π for the Dirichlet kernel Dk (t) =   2N+1 2π 1  1 sin((N+t 2 )t ) 2π sin 2 t=0 otherwise. We can use this fact to find that N σN f (x) = 1 X Sk f (x) = N +1 k=0 Z N π −π 1 X Dk (x − t)f (t)dt, N +1 k=0 and thus we know that the desired kernel is N KN (x − t) = 1 X Dk (x − t). N +1 k=0 We can now substitute in our expression for Dk , using the variable x instead of x − t. The case x = 0 can be done easily (we just have constants), and for all other x we can slightly rewrite our expression as    N X 1 1 x 1 KN (x) = 2 sin sin k+ x .
2π(N + 1) 2 sin x 2 2 2 k=0 2 By the trig product-to-sum identity, this simplifies to = N X 1 1 cos(kx) − cos ((k + 1)x) ,
2π(N + 1) 2 sin x 2 k=0 2 and this is a telescoping sum which simplifies to 1 1
(1 − cos((N + 1)x)) . 2π(N + 1) 2 sin x 2 2
1−cos x x We can now use another trig formula = cos2 2 to get 2   1 1 N +1 2 = x ,
sin 2π(N + 1) sin x 2 2 2 = which is indeed the expression for our Fejér kernel. 84 We can now verify the properties of the Fejér kernel directly: (1) is true because we have a manifestly positive expression and sin2 (cx) is even, and (2) is true because sin2 is also periodic with half the period of the corresponding sin. For (3), notice that Z π Z k X π Dk (t)dt = −π e int dt, −π n=−k and the integral of e int is zero unless n = 0 (by 2π-periodicity), so we just pick up the n = 0 term and get 1. Since σN is the average of the Dk s, the integral of σN is also the average of the average of the Dk s, which will also be 1. Finally, for (4), notice that sin2 x 2 is an even function which is increasing on [0, π]. So if we pick some δ ∈ (0, π), we can say that δ ≤ |x| ≤ π =⇒ sin2 x δ ≥ sin2 , 2 2 so we indeed get the expected 1 KN (x) = |KN (x)| ≤ 2π(N + 1) sin2 2 δ 2  sin N +1 x 2  ≤ 1 . 2π(N + 1) sin2 2δ Now, we can prove convergence of the Cesaro means σN f to f by first doing it for continuous functions – we showed that the continuous functions with endpoints 0 are dense in L2 (so we can show convergence appropriately), and continuous functions with endpoints both 0 can indeed be treated as 2π-periodic. So the subspace of 2π-periodic continuous functions is dense in L2 , and we’ll consider this dense subset first because it’s where the heuristic argument we made above applies rigorously. Theorem 174 (Fejér) Let f ∈ C([−π, π]) be 2π-periodic (so f (−π) = f (π)). Then σN f → f uniformly on [−π, π]. In other words, we have an even stronger result than L2 convergence, now that we’re limiting ourselves to continuous functions and have the stronger uniform norm. But this does not imply that the Fourier series of f converges pointwise to f – there are indeed Fourier series representations of continuous functions that diverge at a point. Instead, it’s the Cesaro mean and the Fejér kernel that help us out here! Proof. First, we extend f to all of R by periodicity (defining it so that f (x + 2π) = f (x) for all x ∈ R). Our function is then an element of C(R) (still continuous), and it is 2π-periodic, so it is uniformly continuous and bounded on all of R (that is, ||f ||∞ = supx∈[−π,π] f (x) < ∞). We wish to show that σN f converge uniformly on f , which means that for all ε > 0 we need to find an M so that for all n ≥ M, we have |σN f (x) − f (x)| < ε for all x. Indeed, for any ε > 0, by uniform continuity of f , there exists some δ > 0 so that for all |y − z| < δ, we have |f (y ) − f (z)| < N ≥ M, we have 2||f ||∞ (N + 1) sin2 δ 2 < ε 2. So now we can choose M ∈ N so that for all ε . 2 (we can do this because the left-hand side converges to 0 as N → ∞). Now because f and KN are 2π-periodic, we can write the Cesaro mean as Z π Z x+π KN (x − t)f (t)dt = σN f (x) = −π KN (τ )f (x − τ )dτ x−π by a change of variables (which is allowed because we’re doing integrals over continuous functions, and thus we can use the Riemann integral), and now we have the product of 2π-periodic functions, so the integral of that is the same 85 over any interval of length 2π: switching back to t from τ , Z π = KN (t)f (x − t)dt. −π We can now say that for all N ≥ M and for all x ∈ [π, π], we have Z π Z |σN f (x) − f (x)| = KN (t)f (x − t)dt − −π where we’ve added in a Rπ −π π KN (t)f (x)dt −π KN (t)dt integral to the f (x) term, which is okay because f (x) doesn’t talk to the t-integral. Combining the integrals by linearity gives us Z π = KN (t) (f (x − t) − f (x)) dt . −π We’ll use the triangle inequality and then split this integral into two parts now: Z π Z Z ≤ |KN (t) (f (x − t) − f (x))| dt = KN (t) |f (x − t) − f (x)| dt + −π |t|≤δ KN (t) |f (x − t) − f (x)| dt δ≤|t|≤π (also using the fact that KN is always nonnegative). And now we can use our bounds above to simplify this: for the first term, we know that |(x − t) − x| < δ over the bounds of integration, so |f (x − t) − f (x)| < 2ε . And for the second term, we know that |f (x − t) − f (x)| < 2||f ||∞ because both f (x − t) and f (x) have magnitude at most ||f ||∞ for a continuous function, and when |t| > δ we can use condition (4) of the Fejér kernel. Putting this all together, we find the inequality ε < 2 2||f ||∞ KN (t)dt + 2π(N + 1) sin2 |t|≤δ Z Z δ 2 KN (t)dt. δ≤|t|≤π We can now bound both integrals here by the integral over the entire region to get ≤ ε 2||f ||∞ + 2 (N + 1) sin2 δ 2 < ε ε + =ε 2 2 by our choice of N. So we’ve indeed shown uniform convergence – σN f is eventually close enough to f for large enough N – and we’re done. Remark 175. This same proof can be modified if instead of knowing that Kn (x) ≥ 0 (which we know for the Fejér kernel), we have that Z π |KN (x)| < ∞. sup N −π Then we can show the same uniform convergence by modifying our proof above. But if we try to plug in our Dirichlet kernel here, the condition is not satisfied, since Z π |DN (x)|dx ∼ log N. −π So having “almost all of the properties” isn’t enough for us to get the analogous results for the Dirichlet kernel! Now that we’ve proven that the Cesaro means of a continuous function converge uniformly to that function, we want to show that the Cesaro means of an L2 function converge to an L2 function, which would show the condition on the Hilbert space that we want and show convergence of the Fourier series as well. We’ll first need the following result: 86 Proposition 176 For all f ∈ L2 ([−π, π]), we have ||σN f ||2 ≤ ||f ||2 . Proof. We’ll first do this for 2π-periodic functions. First suppose that f ∈ C([−π, π]) is 2π-periodic – extend f to all Rπ of R as before, and then the Cesaro mean is σN f (x) = −π f (x − t)KN (t)dt. Thus, we can write out ||σN f ||22 = Z π |σN f (x)|2 dx = Z π Z π Z π f (x − s)f (x − t)KN (s)KN (t)dsdtdx. −π −π −π −π All of these functions are continuous, so we can change the order of integration by Fubini’s theorem to get  Z π Z πZ π = KN (s)KN (t) f (x − s)f (x − t)dx dsdt. −π −π −π By Cauchy-Schwarz, this can be bounded by Z πZ π ≤ KN (s)KN (t)||f (· − s)||2 ||f (· − t)||2 dsdt, −π −π where f (· − s) denotes the function that maps x 7→ f (x − s). And now we’re integrating a periodic function f (· − s) over an interval of length 2π, so we can replace that expression with ||f ||2 (just shifting to another length 2π interval). Doing the same with f (· − t) gives us = ||f ||22 Z π −π Z π KN (s)KN (t)dsdt = ||f ||22 , −π because the integral of KN is 1. This gives us the desired inequality for 2π-periodic functions, and now to extend it to all functions in L2 , suppose we have some general f ∈ L2 . From exercises, we know that there exists a sequence {fn }n of 2π-periodic continuous functions that converge to f in L2 , meaning that ||fn − f ||2 → 0. So from the definition of the Cesaro means, this means that ||σN fn − σN f ||2 → 0 for any fixed N and as N → ∞, leading us to ||σN f ||2 = lim ||σN fn ||2 ≤ lim ||fn ||2 n→∞ n→∞ (using the 2π-periodic case), and this last result is ||f ||2 because fn converges to f in L2 . So now we’re almost done, and combining the two results above will give us what we want: Theorem 177 For all f ∈ L2 , ||σN f − f ||2 → 0 as N → ∞. Therefore, if fˆ(n) = 0 for all n, then f = 0 (since σN f = 0 for all N). Proof. (We only need to prove the result in the first sentence – the second follows directly as stated.) Let f ∈ L2 ([−π, π]), and let ε > 0. By density of the 2π-periodic continuous functions, there exists some 2π-periodic g ∈ C([−π, π]) so that ||f − g||2 < 3ε . Because σN g → g uniformly on [−π, π], there exists some M so that for all N ≥ M and for all x ∈ [−π, π], we have |σN g(x) − g(x)| < √ε . 3 2π Now for all N ≥ M, the triangle inequality tells us that ||σN f − f ||2 ≤ ||σN f − σN g||2 + ||σN g − g||2 + ||g − f ||2 . The first term is ||σN (f − g)||2 (we can check this from the definition), and by Proposition 176, that is less than
1/2 Rπ ||f − g||2 < 3ε . Meanwhile, the last term is also bounded by 3ε , and the middle term is −π |σN g(x) − g(x)|2 dx < 87  2π ·  √ε 3 2π 2 1/2 = 3ε . So putting this all back into our expression gives us ||σN f − f ||2 < ε ε ε + + = ε, 3 3 3 completing the proof. So we’ve now seen a concrete application of the general machinery we’ve built up for Hilbert spaces! In summary, we’ve shown that the normalized exponentials form a maximal orthonormal set, so that the partial Fourier sums of f converge to f in L2 . But as previous mentioned, we don’t have pointwise convergence everywhere – instead, we can only say that there is a subsequence that converges to f pointwise. And in fact, Carleson’s theorem is a deep result in analysis that tells us that for all f ∈ L2 , SN f (x) → f (x) almost everywhere. We can also ask questions about the convergence of Fourier series in other Lp spaces, since all of the definitions also make sense there. It is known additionally that for all 1 < p < ∞, we always have ||SN f − f ||p → 0, and that this is false for p = 1, ∞. But deeper harmonic analysis is needed to prove statements like this, and in particular we would need to learn how to work with singular integral operators. In this class, though, this is as far as we’ll go with Fourier series, and next time, we’ll move on to the topic of minimizers over closed convex sets and (as a consequence) how to identify the dual of a Hilbert space with the Hilbert space itself in a canonical way. 88