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We may replace with another sequence so that all the are continuous and differentiable and are periodic.
Let be a linear, piecewise approximation of ( and have the same values at the endpoints of the interval). Because is piecewise continuous, the square integral of the approximation will also approximate the square integral of . The “edges” of may be smoothed out by composing with functions of the form while retaining its properties of approximation.
Let be the following subset of the set of square integrable functions :
We claim that every function has a weak derivative. To do this, suppose that is a function which is an limit of a sequence . Since , then we may show that the sequence converges weakly to a limit . To do so, observe that if is a sequence of functions in such that is bounded, then there is a weakly convergent subsequence.
The derivatives , are a bounded sequence in .
By Lemma 1, for a finite . Recall the definition of . Because is the integral finite sum of these positive quantities, these positive quantities must in turn be finite and thus, bounded.
Then, because these derivatives , are a bounded sequence in , they have corresponding subsequences which converge weakly to functions .
Hence, we can define the weak energy for any function . Let
then . We will show that this inf is attained.
Take a sequence so that Here we prove that a subsequence have an limit
For this, note that by definition of we can choose for each a smooth function so that
There is a continuous function such that converge uniformly to .
Hence, the sequence converges uniformly to some function . By the triangle inequality, in , as well. Since is a closed set in , then and accordingly, has a weak derivative .
The sequence converges weakly to .
We take advantage of the fact that for every , there is a sequence of test functions such that in .
So then, take any We need to show that
For this, take using the triangle inequality:
and using Cauchy-Schwartz:
using the definition of the sequence
and using the definition of the weak derivative, since is a smooth test function:
and using Cauchy-Schwartz again we have:
Letting FIRST, and then we get the result.
By Lemma 3, since weakly, we get
We may show that . Because is continuous, as ; thus, because by definition of weak convergence, .
We have shown that the components and allow to attain the minimum . Now, note that if is a test function, then and so we automatically get
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