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Lemma 2

We may replace V k with another sequence so that all the ϕ k are continuous and differentiable and are 2 π - periodic.

Proof:

Let ϕ 1 be a linear, piecewise approximation of ϕ ( ϕ and ϕ 1 have the same values at the endpoints of the interval). Because ϕ is piecewise continuous, the square integral of the approximation ϕ 1 will also approximate the square integral of ϕ . The “edges” of ϕ 1 may be smoothed out by composing ϕ 1 with functions of the form e 1 x while retaining its properties of approximation.

Let H be the following subset of the set of square integrable functions L 2 :

H = ϕ L 2 : there is a sequence ϕ k C such that | ϕ - ϕ k | 2 0 .

We claim that every function ϕ H has a weak derivative. To do this, suppose that ϕ is a function which is an L 2 limit of a sequence ϕ k C . Since E ( ϕ ) < E + 1 , then we may show that the sequence ( ϕ k ) t converges weakly to a limit ϕ t . To do so, observe that if f k is a sequence of functions in L 2 such that | | f k | | L 2 is bounded, then there is a weakly convergent subsequence.

Lemma 3

The derivatives ϕ k , θ , ϕ k , t are a bounded sequence in L 2 .

Proof:

By Lemma 1, E ( ϕ k ) E ( ϕ ) = E for a finite E . Recall the definition of E = c o s 2 ϕ + ϕ t 2 + ϕ θ 2 d t d θ . Because E is the integral finite sum of these positive quantities, these positive quantities must in turn be finite and thus, bounded.

Then, because these derivatives ϕ k , θ , ϕ k , t are a bounded sequence in L 2 , they have corresponding subsequences which converge weakly to functions ϕ t , ϕ θ .

Hence, we can define the weak energy E ( ϕ ) for any function ϕ H . Let

E H = inf ϕ H E ( ϕ ) ,

then E H E . We will show that this inf is attained.

Take a sequence ϕ k H so that E ( ϕ k ) E H . Here we prove that a subsequence ϕ k have an L 2 limit ϕ H .

For this, note that by definition of H we can choose for each ϕ k a smooth function ϕ k C so that

ϕ k - ϕ k L 2 < 1 k .

Lemma 4

There is a continuous function ϕ such that ϕ k converge uniformly to ϕ .

Proof:

| ϕ k ( x ) - ϕ k ( y ) | = y x ϕ k ' ( t ) d t y x d t 1 2 y x ϕ ' k 2 ( t ) 1 2
= | x - y | 1 2 ϕ ' k 2 ( t ) 1 2 | x - y | ( E + 1 ) 2 π

Hence, the sequence ϕ k converges uniformly to some function ϕ . By the triangle inequality, ϕ k ϕ in L 2 , as well. Since H is a closed set in L 2 , then ϕ H and accordingly, ϕ has a weak derivative ϕ t .

Lemma 5

The sequence ( ϕ k ) t converges weakly to ϕ t .

Proof:

We take advantage of the fact that for every f L 2 , there is a sequence of test functions f m C c ( ( 0 , 1 ) ) such that f m f in L 2 .

So then, take any f L 2 . We need to show that

( ϕ k ) t f - ϕ t f 0 .

For this, take using the triangle inequality:

( ϕ k ) t f - ϕ t f | ( ϕ k ) t | | f - f m | + ( ( ϕ k ) t - ϕ t ) f m + | ϕ t | | f - f m |

and using Cauchy-Schwartz:

( ϕ k ) t L 2 f - f m L 2 + ( ( ϕ k ) t - ϕ t ) f m + ϕ t L 2 f - f m L 2

using the definition of the sequence ϕ k :

( E H + 1 ) f - f m L 2 + ( ( ϕ k ) t - ϕ t ) f m + ϕ t L 2 f - f m L 2

and using the definition of the weak derivative, since f m is a smooth test function:

= ( E H + 1 ) f - f m L 2 + ( ϕ k - ϕ ) ( f m ) t + ϕ t L 2 f - f m L 2

and using Cauchy-Schwartz again we have:

( ϕ k ) t f - ϕ t f
( E H + 1 ) f - f m L 2 + ϕ k - ϕ L 2 ( f m ) t L 2 + ϕ t L 2 f - f m L 2 .

Letting k FIRST, and then m , we get the result.

By Lemma 3, since ( ϕ k ) t ϕ t weakly, we get

ϕ t 2 lim k ( ϕ k ) t 2 .

We may show that c o s 2 ϕ k c o s 2 ϕ . Because f ( x ) = c o s 2 ( x ) is continuous, c o s 2 ( x ) c o s ( y ) as | x - y | 0 ; thus, because ϕ k - ϕ 0 by definition of weak convergence, c o s 2 ϕ k c o s 2 ϕ .

We have shown that the components ϕ t 2 and c o s 2 ϕ allow ϕ to attain the minimum E H . Now, note that if η is a test function, then ϕ + η H , and so we automatically get

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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