Lecture
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In the metric space, the concept of convergence of a sequence is introduced . Let (X, d ) be a metric space.
Definition 1. They say that x n Î X converges to x Î X (x n ® x; ) if d (x n , x)
® 0 as n ® ¥ .
The concept of convergence can also be formulated in the language " e - n ". For " e > 0 $ n 0 ( e ): for " n ³ n 0 the inequality d (x n , x) < e is true .
Lemma 1. If a sequence converges in a metric space, then its limit is unique.
Evidence. Let x n ® a, x n ® b . Applying the triangle inequality, we get: d ( a, b ) £ d ( a, x n ) + d ( x n , b ). Both terms on the right side tend to zero. Since d ( a, b ) non-negative and is independent of n , then the well-known theorems about the transition to the limit in the inequalities obtain d ( a, b ) = 0, and then the properties of the metric and = b , as required to prove.
Definition 2. A sequence x n of elements of a metric space X is called bounded if there is a ball S ( y , r ) to which all the members of the sequence belong.
Lemma 2. If a sequence converges in a metric space, then it is bounded.
Evidence. The assertion easily follows from the definition of a convergent sequence, if we note that if x n ® x, then for " fixed e > 0 there is n 0 for which x n Î S ( x , e ) for all n ³ n 0. Consequently, all the members of the sequence, with the exception of a finite number, fall into the neighborhood of S ( x , e ) Since any finite set of elements is always bounded, this already implies the boundedness of the entire sequence.
Consequence. If a sequence {x n } of points from X converges to a point x Î X, then the numbers d (x n , y) are bounded for any fixed point in the space X.
Lemma 3. If x n → x, y n → y, then d (x n , y n ) → d (x, y) (in other words, the metric is a continuous function of its arguments).
Evidence. By the quadrilateral inequality
| d (x, y) - d (x n , y n ) |
d (x, x n ) + d (y, y n ).
Hence, passing to the limit as n → we easily obtain the assertion of the lemma.
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Play gameLet us clarify the specific meaning of convergence in the metric spaces R n , C [a, b], l 2 and m.
Example 1. Let X = R n . If x k → x 0 , where x k = {ξ 1 (k) , ..., ξ n (k) } and x 0 = {ξ 1 (0) , ..., ξ n (0) }, then
d (x k , x 0 ) = → 0 as k → ∞.
But by virtue of easily verified inequalities
,
valid for any i , this is possible if and only if ξ i ( k ) → ξ i (0) , i = 1, ..., n, as k → ∞.
Hence it follows that convergence in R n is the convergence of the coordinates of the points of the sequence to the corresponding coordinates of the point - limit, i.e. convergence in R n is convergence in coordinates .
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Play gamed (x n , x 0 ) = | x n (t) - x 0 (t) | → 0
or otherwise: ε> 0 N: n> N =>
| x n (t) - x 0 (t) | <ε. This condition is equivalent to the condition that n> N => | x n (t) - x 0 (t) | <ε t [ a , b ]. But this means uniform convergence of the sequence {x n (t)} to x 0 (t). Thus, the convergence in the space С [ a , b ] is the uniform convergence of the functional sequence {x n (t)}.
Example 3. Let X = l 2 . It can be shown that the convergence of the sequence {x n } l 2 to x 0 l 2 , where x n = {ξ i (n) }, x 0 = {ξ i (0) } means that
1) ξ i (n) → ξ i (0) for i = 1,2, ...
2) ε> 0 N: <ε for all n = 1,2, .....
Thus, convergence in l 2 contains stronger requirements than convergence in coordinates. Let us show this with an example showing that in l 2 the convergence in coordinates does not imply the convergence of a sequence of points in l 2 .
In the space l 2 we take the sequence e m = {ξ i (m) }, where ξ i (m) = δ mi (the Kronecker symbol). We take x 0 = (0, 0,…, 0,…) l 2 . Then the sequence {e m } in coordinates tends to the point x 0 . However,
d (e m , x 0 ) = 1, therefore {e m } does not tend to x 0 in the metric.
Example 4. Let X = m . The convergence of the sequence x n = {ξ 1 (n) ,…, ξ n (n) ,…} m to the element x 0 = {ξ 1 (0) ,…, ξ n (0) ,…} means uniform convergence in coordinates , i.e.
" ε> 0 N:
" n> N Þ | ξ i ( n ) - ξ i (0) | <ε " i = 1,2, ... This is proved in the same way as in Example 2.
It can be shown that in the metric space s of all numerical sequences the convergence in the metric coincides with the convergence in the coordinates.
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Play gameLemma 4 (on the convergence of sequences ). Let { x n } be a sequence from the metric space X. The following conditions are equivalent:
1. { x n } - converges to x ;
2. Any subsequence { x n } converges to x ;
3. For any subsequence { } there is a subsequence { } converging to x;
4 . { x n } is fundamental and any subsequence { } converges to x ;
5. { x n } is fundamental and there is a subsequence { } converging to x .
Evidence.
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Play game4. Þ 5. Obvious.
3. Þ 4. follows from 5. Þ 1. Indeed, if 5. Þ 1. has already been proved, then by the conditions of item 4. the subsequence { } is fundamental, but by item 3 it has a subsequence converging to x . Then from 5.
Þ 1. it follows that { } itself converges to x .
5. Þ 1. Let { x n } be a fundamental sequence and be its subsequence converging to x . For
" e > 0 $ N 1 : d ( x p , x m ) < e , p, m> N 1. Putting here m = n k , n k ³ N 1 , k ³ N, we have d ( x p , ) < e . Consequently, d ( x , x p ) £
d ( x , ) + d ( ,
x p ) £ e + e £ 2 e (p> N 1 ) and x p ® x Î H.
Definition 4. A metric space X is called complete if any fundamental sequence in this space converges to an element of this space.
Example 5. For the case of R n - Euclidean n-dimensional space - the completeness follows from the Cauchy criterion for the existence of the limit of the sequence of points of this space.
Example 6. Consider the space C [0, 1] introduced above. By the definition of the fundamental sequence { x n } and the metric for " e > 0 $ N: < e " n, m ³ N. If we fix t, then x n (t) will be an ordinary numerical fundamental sequence, which exists due to the criterion Cauchy is the pointwise limit x ( t ). Passing to the pointwise limit in the inequality, which is true for any t Î [0, 1] as m ® ¥, we obtain £ e for " n ³
Of N . Thus, the sequence x n ( t ) converges uniformly on the segment [0, 1] to the function x ( t ). Then, by the Weierstrass theorem on the continuity of the uniform limit of continuous functions, x (t) is a continuous function on the interval [0, 1]. Hence C [0, 1] is a complete space.
Example 7. On the set C [0, 1], you can introduce another metric, for example:
d (x, y) =
but in this case the space will not be complete. To prove this, it suffices to consider the following sequence of continuous functions:
x n ( t ) =
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Play gameExample 8. Let us show the completeness of the space l 2 . Let the sequence x ( m ) = ( x 1 ( m ) , x 2 ( m ) , ..., x n ( m ) , ....), m = 1, 2, .... is fundamental in l 2 . Therefore, for an arbitrarily chosen e > 0, there exists a number n 0 such that for all k , m ³ n 0 , the inequality <
e . From the inequality | x n ( m ) - x n ( k ) | £ , valid for any n Î N , it follows that the sequence { x n ( m ) } is fundamental in the space R and hence its convergence x n ( m ) ® x n as m ® ¥ . Moving on in the obvious inequality
< e
for a fixed m to the limit first as k ® ¥ , then as p ® ¥ , we obtain the inequality
£ e .
From the triangle inequality
it follows that x belongs to l 2 . The previous inequality implies the convergence of x ( m ) to x in l 2 .
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Functional analysis
Terms: Functional analysis