24. First basic lemma.

If function g(t) is integrable in some finite interval [a,b] then and, analogically,

Proof. It is sufficiently to prove only first expression. At first let’s notice that for any finite interval we have estimation (5). Let’s divide interval [a,b] into n parts by inserting points(6) and according to that also divide our integral

Denoting as m_i exact low bound of values g(t) in i-th interval, the last expression may be transformed as

Ifis oscillation of function g(t) in i-th interval then inequalitytakes place within this interval; now it’s easy to get estimation of our integral taking into account (5):

Let’s take arbitrary; then we must choose dividing (6) so that it is executable owing to the integrability of function g(t). Now we can take because numbers m_i are defined yet. And we have got for these values q.e.d. If we remember formulas expressed Fourier coefficients than the first corollary will be next. Fourier coefficients a_m, b_m of absolutely integration function approaches to zero, when .

]25. Principle of localization.

The second corollary is the principle of localization.

Let’s take arbitrary positive number and divide integral

into two parts:If second integral will be rewrote as:then it is clear, thatmultiplier at sine is absolutely integration function of t in interval because thedenominator in this interval doesn’t equal to zero. In that case according to lemma this integralapproaches to zero, when. So existence of limit for partial sum of Fourier seriesand value of this limit are a depend of behavior of just only one integral(7)

But this integral contains only function values corresponding to argument’s changing in interval from toThis fact provesprinciple of localization that means next:

Riemann’s theorem.

Behavior of Fourier series of function f(x) in some point x₀ depends only of function values in immediate proximity in relation to considered point.

So, for example, if we take two functions, values of these in arbitrary small neighborhood of point x₀ coincides than corresponding Fourier series in point x₀ behave analogically: or both converges to the sum or both diverge.

26. Dini test of Fourier series’ convergence

Dini test. Fourier series of function f(x) in point x₀ converges to sum S₀ if for some h>0 integral exists.

Proof. Let’s rewrite (8) as. Taking into account basic lemma we can see that whenit approaches to zero because function and together with it is absolutely integrated. Then integral exist, therefore exists, q.e.d.

27. Lipschitz test test of Fourier series’ convergence

Fourier series of continuous function in the point converges to sum if for sufficiently small t takes place, where L and positive constants (). Proof. Ifthen, so integrals (11): ,exist.

If then, at the right side we have integrated function therefore integrals (11) exist too.

Condition of Lipschitz test when will be executed if function has finite derivative in point x₀ or one-sided finite derivatives and , though its differ one from another. So, in point x₀ where function f(x) is differentiable or has both finite one-side derivatives Fourier series converges to f(x₀).

Lipschitz test may be easily formulated for case when. In point x₀ where function has first type jump for convergence of Fourier series it’s sufficiently to suppose existence of finite limits

and , and sum of series will be . So, if we have with period and differentiated or piecewise-differentiable then its Fourier series always converges to in point excluding jump points, where sum will be .

28. Dirichlet test. If function f(x) with period 2π is picewise-monotonuos in interval and has in it finite number of jumps then its Fourier series converges to sum f(x₀) in every continuosity point and to sum in every jump point.

Proof is based on that fact that monotonous and bounded function g(t) satisfies to condition where h>0, the last expression we take without proof. Conditions mentioned in the last test are called Dirichlet’s condition and are used often than others tests.