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FREDHOLM OPERATORS (NOTES for FUNCTIONAL ANALYSIS) We Assume Throughout the Following That X and Y Are Banach Spaces Over the Fi

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FREDHOLM OPERATORS (NOTES for FUNCTIONAL ANALYSIS) We Assume Throughout the Following That X and Y Are Banach Spaces Over the Fi FREDHOLM OPERATORS (NOTES FOR FUNCTIONAL ANALYSIS) CHRIS WENDL We assume throughout the following that X and Y are Banach spaces over the field K P tR; Cu unless otherwise specified. For a bounded linear operator T P L pX; Y q, we denote its transpose, also known as its dual operator, by T ˚ : Y ˚ Ñ X˚ :Λ ÞÑ Λ ˝ T . Contents 1. Definitions and examples 1 1 ´1 2. The Sobolev spaces H0 pΩq and H pΩq 2 3. Main theorems 6 4. Some preparatory results 9 5. The semi-Fredholm property 11 6. Compact perturbations 12 References 14 1. Definitions and examples A bounded linear operator T : X Ñ Y is called a Fredholm operator if dim ker T ă 8 and dim coker T ă 8; where by definition the cokernel of T is coker T :“ Y im T; so the second condition means that the image ofLT has finite codimension. The Fredholm index of T is then the integer indpT q :“ dim ker T ´ dim coker T P Z: Fredholm operators arise naturally in the study of linear PDEs, in particular as certain types of differential operators for functions on compact domains (often with suitable boundary conditions imposed). 1 Example 1.1. For periodic functions of one variable x P S “ R{Z with values in a finite- k 1 k´1 1 dimensional vector space V , the derivative Bx : C pS q Ñ C pS q is a Fredholm operator with index 0 for any k P N. Indeed, 1 k 1 ker Bx “ constant functions S Ñ V Ă C pS q; and ( k´1 1 im Bx “ g P C pS q gpxq dx “ 0 ; 1 " ˇ »S * the latter follows from the fundamental theorem ofˇ calculus since the condition g x dx 0 ˇ S1 p q “ ensures that the function fpxq :“ x gptq dt on isˇ periodic. The surjective linear map 0 R ³ Ck³´1pS1q Ñ V : g ÞÑ gpxq dx 1 »S thus has im Bx as its kernel, so it descends to an isomorphism coker Bx Ñ V , implying indpBxq “ dim V ´ dim V “ 0. 1 2 CHRIS WENDL k,α 1 k´1,α 1 Example 1.2. For the same reasons as explained in Example 1.1, Bx : C pS q Ñ C pS q is Fredholm with index 0 for every k P N and α P p0; 1s. 8 1 8 1 Exercise 1.3. Use Fourier series to show that the unique extension of Bx : C pS q Ñ C pS q to a bounded linear operator Hs`1pS1q Ñ HspS1q is also Fredholm with index 0 for every s ¥ 0. Exercise 1.4. Show that for functions taking values in a vector space V of dimension n, the k k´1 derivative Bx : C pr0; 1sq Ñ C pr0; 1sq is a surjective Fredholm operator with index n, but imposing the boundary condition fp0q “ fp1q “ 0 produces an injective Fredholm operator f P Ckpr0; 1sq fp0q “ fp1q “ 0 ÝÑBx Ck´1pr0; 1sq ! ˇ ) with index ´n. ˇ ˇ Exercise 1.5. Show that for n ¥ 2 and each j “ 1; : : : ; n, the bounded linear operators k n k´1 n k,α n k´1,α n s`1 n s n Bj : C pT q Ñ C pT q, Bj : C pT q Ñ C pT q and Bj : H pT q Ñ H pT q have infinite-dimensional kernels and are thus not Fredholm. n 2 Example 1.6. The Laplacian ∆ :“ j“1 Bj on fully periodic functions of n variables valued in a finite-dimensional vector space V defines a Fredholm operator ř s`2 n s n ∆ : H pT q Ñ H pT q s 2 n with index 0 for each s ¥ 0. Indeed, if u P H ` pT q and f “ ∆u, then f has Fourier coefficients n f “ B2u “ ´4π2|k|2u P V; k j k k j“1 ¸ thus p y p 2 2 s 2 4 2 s 4 2 4 2 s`2 2 }f}Hs “ p1 ` |k| q |fk| “ 16π p1 ` |k| q |k| |uk| ¤ 16π p1 ` |k| q |uk| kPZn kPZn kPZn ¸ 4 2 ¸ ¸ “ 16π }u}Hs`2 ; p p p s`2 s n proving that ∆ is bounded from H to H . If ∆u “ 0, then fk “ 0 for all k P Z , implying n uk “ 0 for all k P Z zt0u, but there is no condition on the coefficient u0 P V , thus ker ∆ is the s`2 n space of functions in H pT q whose only nonvanishing Fourierp coefficient is u0, also known n asp the constant functions T Ñ V . Similarly, the equation ∆u “ f canp be solved for a given s n 1 n f P H pT q by writing uk “ ´ 4π2|k|2 fk for all k P Z zt0u, but this is only possiblep if f0 “ 0, thus s n s n im ∆ “ f P H pT q f0 “ 0 , and the surjective linear map H pT q Ñ V : f ÞÑ f0 therefore p p p descends! to an isomorphismˇ coker) ∆ Ñ V . We conclude ind ∆ “ dim V ´ dim V “ 0. ˇ p 2 2 s`2 2 s 2 p Exercise 1.7. Show that the wave operator Bt ´ Bx : H pT q Ñ H pT q for fully periodic 2 functions of two variables pt; xq P R has infinite-dimensional kernel, so it is not Fredholm. Hint: Consider functions of the form pt; xq ÞÑ fpt ˘ xq. n Example 1.8. On any bounded open domain Ω Ă R , the Laplacian defines bounded linear operators Ck`2pΩq Ñ CkpΩq, Ck`2,αpΩq Ñ Ck,αpΩq for each k ¥ 0 and α P p0; 1s, as well as W k`2;ppΩq Ñ W k;2pΩq for each p P r1; 8s, but none of these operators are Fredholm. The n reason is that alls smooth solutionss to the equation ∆u “ 0 on R (these are called harmonic functions) belong to the kernels of these operators, and there is an infinite-dimensional space of 2 such solutions. This is especially easy to see in the case n “ 2, where one can identify R “ C and extract harmonic functions from the real parts of holomorphic functions C Ñ C. 1 ´1 2. The Sobolev spaces H0 pΩq and H pΩq The Laplacian ∆ is the most popular example of an elliptic operator; in contrast to the wave operator of Exercise 1.7, it has the right properties to produce a Fredholm operator in suitable functional-analytic settings, as demonstrated by Example 1.6. The problem with Example 1.8 FREDHOLM OPERATORS 3 turns out to be not the operator ∆ itself, but the fact that it is being considered on a bounded domain without imposing any boundary condition.1 To discuss the Laplacian with boundary conditions, it is useful to introduce a few new varia- s n tions on the usual Sobolev spaces H pR q. We shall assume in the following that all functions take values in a fixed finite-dimensional complex inner product space pV; x ; yq unless otherwise s n 2 n noted. Recall that H pR q is defined for each s ¥ 0 as the space of functions f P L pR q n with the property that the product of the Fourier transform f : R Ñ V with the function n 2 s 2 2 n R Ñ R : p ÞÑ p1 ` |p| q { is also in L pR q. The same definition does not quite make sense for s ă 0 since in that case, p1 ` |p|2qs{2f could very well be ofp class L2 without f itself being 2 s n 2 n of class L , in which case one should not require H pR q to be a subspace of L pR q. The s n remedy is to define H pR q for s ă 0 as ap space of tempered distributions rather thanp functions. In fact, the resulting definition also makes sense for s ¥ 0, but reduces then to the previous definition since tempered distributions whose Fourier transforms are L2-functions can always be represented by L2-functions. s n n Definition 2.1. For any s P R, we define H pR q Ă S 1pR q as the space of all tempered distributions Λ whose Fourier transforms are represented by functions of the form Λppq “ p1 ` 2 s 2 2 n s |p| q´ { fppq for some f P L pR q. The H -norm is then defined via the inner product p 1 2 s{2 2 s{2 1 2 s 1 xΛ; Λ yHs :“ p1 ` |p| q Λ; p1 ` |p| q Λ “ p1 ` |p| q xΛppq; Λ ppqy dp: L2 n »R A s n E It is easy to see that H pR q isp a Hilbert space,p as it admits a naturalp unitaryp isomorphism 2 n 2 s 2 to L pR q, defined by taking Fourier transforms and multiplying by p1 ` |p| q { . Exercise 2.2. Assuming s P R, prove: n s n (a) A distribution Λ P S 1pR q is in H´ pR q if and only if it satisfies a bound |Λp'q| ¤ n c}'}Hs for all test functions ' P S pR q. n s n (b) The space of vector-valued Schwartz-class functions S pR q is dense in H pR q. n n (c) The pairing S pR q ˆ S pR q Ñ C : p'; q ÞÑ x'; yL2 extends to a continuous real- bilinear pairing ´s n s n x ; ys : H pR q ˆ H pR q Ñ C; 2 ´s{2 2 s{2 xΛ; fys :“ p1 ` |p| q Λ; p1 ` |p| q f “ xΛppq; fppqy dp; L2 n R A E ´s» n such that the real-linear map Λ ÞÑ xΛ; ¨ys sends H pR q isomorphically to the dual s n p p p p space of H pR q.
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