STUDY GUIDE FOR BASIC EXAM

STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There re surely errors nd typos in this document, so plese use these notes crefully. If you do hppen to find n error or typo, feel free to e-mil me. 1. Anlysis 3.1 Prove tht C(X is complete for ny metric spce X. 3.2 Prove tht if f n f uniformly on the metric spce X where ech f n is continuous, then f is 3.3 Prove tht f n f uniformly on [, b] implies f n f. 3.4 Uniform continuity; extension theorem for uniformly continuous functions. 3.5 Prove tht continuous function is Riemnn integrble. 3.6 Dini s Theorem. 3.7 Proof of chin rule for functions f : R n R m. Implicit function theorem. G δ nd F σ sets; show tht Q is not G δ. Distnce between closed subsets of metric spce. Rerrngement of series. Does (f n continuous imply f n converge? If f n f, does f n f? Prove if... is compct subset of C(X for some compct X (usully [, b]. 2. Liner Algebr 3.8 Rnk-nullity theorem dim rn T + dim ker T = dim V. 3.9 (rn T = ker T 3.10 Digonliztion of symmetric mtrices; Spectrl Theorem. 3.11 Show tht eigenvectors with distinct eigenvlues re linerly independent. Annihiltor subspces. 3. Proofs 3.1. C(X is complete for ny metric spce X. Proof. Let f n be Cuchy sequence of functions in C(X. Since f n is Cuchy, for ech x X the sequence f n (x is Cuchy, hence convergent in C. Define f : X C by f(x = lim n f n (x. We wish to show tht f n f uniformly on X. Let ε > 0. Since f n is Cuchy there is some lrge N so tht m, n N implies f n (x f m (x < ε 2 for ll x X. Letting m while keeping n N fixed, we see tht f n (x f(x = lim n f n(x f m (x ε 2 < ε for ll x X. Tht is, f n f uniformly on X. Furthermore, since ech f n is continuous nd the convergence is uniform, f is Note tht if C is replced with ny complete metric spce Y, the sme proof shows tht C(X, Y is complete. 1

2 BRYON ARAGAM 3.2. If f n f uniformly on the metric spce X where ech f n is continuous, then f is Proof. Let ε > 0 nd choose x 0 X. Since f n f uniformly, there exists N 0 so tht for n N we hve f n (x f(x < ε for ll x X. Moreover, since f N is continuous, there is δ > 0 so tht f N (x f N (x 0 < ε for x x 0 < δ. Thus f(x f(x 0 f(x f N (x + f N (x f N (x 0 + f N (x 0 f(x 0 < 3ε for x x 0 < δ, where we hve used the tringle inequlity severl times. This estblishes tht f is 3.3. If f n f uniformly on [, b] nd f n is Riemnn integrble for ll n, then f is Riemnn integrble nd f n f. Proof. Define ε n := sup { f n (x f(x x b} so tht f n ε n f f n + ε n for ech n. Then (f n ε n dx L[f] U[f] (f n + ε n dx, where L[f] nd U[f] re the lower nd upper integrls of f, respectively. Thus 0 L[f] U[f] 2ε n (b, so s n we see tht L[f] U[f]. Tht is, f is Riemnn integrble. Furthermore, given ε > 0, we hve f n (x f(x < ε for lrge n nd ll x b. This implies f n dx f dx f n f dx ε(b, which implies the theorem. 3.4. Suppose f : A X is uniformly continuous function with A open in the complete metric spce X. There exists n extension of f to the closure of A, tht is, continuous function g : A X such tht g A = f. Proof. Given x A, pick sequence x n A converging to x. Since x n is Cuchy nd f is continuous, it is esy to see tht the sequence f(x n is Cuchy. Since X is complete, this sequence hs limit. Define g(x := lim f(x n. This construction is esily checked to be well-defined: if x n nd x n both tend to x A, then d(x n, x n is rbitrrily smll for lrge n, which by the continuity of f on A implies d(f(x n, f(x n is rbitrrily smll for lrge n. Tht is, f(x n nd f(x n hve the sme limit. It remins to show tht the extension g is continuous on A. Fix ε > 0 nd x A. Choose sequence x n A such tht x n x. Given this sequence, fix sequence x n A tht converges to x nd for ech n, choose sequence x k n A tht converges to x n (which is possible since x n, x A. Now choose N so lrge tht d(g(x N, g(x < ε, which is possible by our definition of g. Now, since g is uniformly continuous on A, we cn find δ > 0 so tht d(x, y < δ implies d(g(x, g(y < ε. Now choose n > N so lrge tht d(x n, x n < δ/2, nd given this n, choose K so lrge tht d(x K n, x n < δ/2. Then we hve d(x K n, x n d(x K n, x n + d(x n, x n < δ so tht d(g(x K n, g(x n < ε. Finlly, choose k > K so tht d(g(x k n, g(x n < ε (gin, possible by the definition of g. Thus d(g(x n, g(x < d(g(x n, g(x k n + d(g(x k n, g(x n + d(g(x n, g(x < 3ε, which shows tht g is

STUDY GUIDE FOR BASIC EXAM 3 3.5. If f : [, b] R is continuous, then f is Riemnn integrble. Proof. Let ε > 0. Since [, b] is compct, f is uniformly continuous on [, b], so there exists δ > 0 so tht x y < δ = f(x f(y < ε b if x, y [, b]. Choose n integer n lrge enough so tht x := b n < δ nd prtition [, b] into subintervls of length x. Cll this prtition P = {x 0,..., x n }. Thus we hve M i m i + ε b by uniform continuity, where Thus M i = sup {f(x x i 1 x x i } m i = inf {f(x x i 1 x x i }. M i x = = ( m i + ε x b m i x + n x ε b m i x + ε, since n x = b by our definition of x. This shows tht the upper integrl of f on [, b] is equl to it s corresponding lower integrl, tht is, f is Riemnn integrble on [, b]. 3.6. (Dini s Theorem Suppose f n f pointwise where f n, f : A R. Then the following conditions imply tht f n f uniformly: i Ech f n is ii The sequence f n is monotonic in n. iii f is iv A is compct. If ny of these conditions fil to hold, then the convergence need not be uniform. Proof. We prove the cse where f n is decresing tht is, f n (x f n+1 (x for ll x A. The incresing cse is similr. Let ε > 0. For ech n, define g n := f n f nd E n := {x A g n (x < ε}. Since ech f n is continuous long with f, g n is continuous, whence E n is open for ech n. Moreover, g n is decresing in n, so we hve E n E n+1 for ll n. Thus the {E n } form n open cover of A. By hypothesis A is compct, so there is finite subcover {E n1,..., E nk } of A. Since the sets E n re incresing, this implies tht E ni = A for some 1 i k. Thus if n > n i, f(x f n (x < ε for ny x A. 3.7. (Chin Rule. If g : R k R m nd f : R m R n re differentible t x 0 R k, g(x 0 R m, respectively, then f g is differentible nd D[f g](x 0 = Df(g(x 0 Dg(x 0. Proof. Since g is differentible t x 0, there exists some M > 0 so tht g(x g(x 0 M x x 0 when x x 0 is sufficiently smll. Let ε > 0. Since f is differentible t g(x 0, we hve f(g(x f(g(x 0 Df(g(x 0 (g(x g(x 0 < ε 2M g(x g(x 0 (1 ε 2 x x 0 for smll x x 0. Note tht we hve used the fct tht g is continuous: When x x 0 is smll, we know tht g(x g(x 0 is smll s well.

4 BRYON ARAGAM We now wish to show tht [Df(g(x 0 Dg(x 0 ] (x x 0 Df(g(x 0 (g(x g(x 0 = Df(g(x0 [g(x 0 Dg(x 0 (x x 0 g(x ] (2 < ε 2 x x 0 when x x 0 is sufficiently smll. But this is cler from the differentibility of g nd the continuity of Df(g(x 0. Combining (1 nd (2 using the tringle inequlity completes the proof. 3.8. For liner trnsformtion T : V W, dim rn T + dim ker T = dim V. Proof. Let {v 1,..., v k } be bsis for ker T. Then we cn extend this collection to bsis of V : {v 1,..., v k, v k+1,..., v n }. We clim tht B = {T v k+1,..., T v n } is bsis for rn T. Given ny element T v of rn T, we hve T v = T α i v i = α i T v i = α i T v i i=k+1 since T v 1 = T v 2 = = T v k = 0. Thus B genertes rn T. Now suppose α k+1 T v k+1 + + α n T v n = T (α k+1 v k+1 + + α n v n = 0. Then α k+1 v k+1 + + α n v n ker T, so α k+1 v k+1 + + α n v n = β 1 v 1 + + β k v k for some sclrs β i, i = 1,..., k. But the collection {v 1,..., v k, v k+1,..., v n } is linerly independent, so β 1 = = β k = α k+1 = = α n = 0. In prticulr, B must be linerly independent. This theorem is lso consequence of the more generl sttement tht given ny vector subspce W V, we hve dim V/W + dim W = dim V. The proof is lmost identicl to the bove. In fct, given ny bsis of W, we cn extend it to bsis for V, nd show tht the new vectors so obtined form bsis for V/W. To connect this fct with the theorem, recll the first isomorphism theorem from bsic lgebr: Given liner trnsformtion T : V W we hve rn T = V/ ker T. In fct, the isomorphism is given by ϕ(v + ker T = T v. The function ϕ is clerly surjective; to prove injectivity note tht if T v = T v then v v ker T, whence v + ker T = v + ker T. Thus dim V = dim(v/ ker T + dim ker T = dim rn T + dim ker T, s desired. 3.9. (Rnk-Nullity Theorem. Given liner trnsformtion T : V W nd it s corresponding dul mp T : W V, (rn T = ker T. Proof. This is just (very creful symbol pushing: w ker T T (w = 0 T (w (v = 0 w (T v = 0 w (rn T. for ll v V for ll v V

STUDY GUIDE FOR BASIC EXAM 5 3.10. Let T : V V be symmetric liner trnsformtion on finite-dimensionl vector spce V (rel or complex. Then V hs n orthonorml bsis consisting of eigenvectors of T. Proof. First we show tht T hs t lest one non-zero eigenvector. If V is C-vector spce, then this is trivil since the chrcteristic polynomil of T must hve t lest one root in C. If V is n R-vector spce, we re not gurnteed tht the root lies in R. Let Q(x = T x, x. Since Q is clerly continuous, it hs mximum Q(p on the unit sphere {v V v = 1} 1. We clim tht p is n eigenvector for Q. To prove this, choose unit vector w W := p nd consider the function g(t = Q(γ(t, where γ(t := (cos tp + (sin tw. Observe tht γ is curve on the unit sphere such tht γ(0 = p nd γ (0 = w. We cn differentite g to obtin g (t = γ (t, T (γ(t + γ(t, T (γ (t = γ (t, T (γ(t + T (γ(t, γ (t = 2 γ (t, T (γ(t since T = T Since Q(p is the mximum of Q, g(0 = Q(p is mximum for g. Thus g (0 = 0, or 0 = 2 γ (0, T (γ(0 = 2 w, T p. Thus T p is perpendiculr to W, tht is, T p W. But W = (p = p, the 1-dimensionl subspce generted by p. Thus T p = λp for some λ R. Thus T hs t lest one eigenvlue v. Let W = v nd note tht if w W, T w, v = λ w, v = 0, so T (W W. Tht is, T W is symmetric liner trnsformtion on the (n 1-dimensionl vector spce W. By induction we cn find n orthogonl bsis of eigenvectors for W. Clerly v must be orthogonl to this bsis, so djoining v yields bsis of mutully orthogonl eigenvectors for V. Normlizing this bsis completes the proof. 3.11. Eigenvectors of liner trnsformtion T with distinct eigenvlues re linerly independent. Proof. We prove this by induction. The clim is trivil for n = 1. Suppose tht {v 1,..., v n } re eigenvectors of T with corresponding eigenvlues λ 1,..., λ n nd α i v i = 0. Then we hve (3 (4 0 = T α i v i = 0 = λ n α i v i = α i T v i = α i λ n v i. α i λ i v i, Subtrcting (4 from (3 we hve n 1 α i (λ i λ n v i = α i (λ i λ n v i = 0. By induction, we cn ssume tht {v 1,..., v n 1 } re linerly independent, so α i (λ i λ n = 0 for i = 1,..., n 1. But λ i λ n for such i by hypothesis, whence α i = 0. This implies α n = 0, proving the theorem. 1 Note tht we re using the fct tht V is finite dimensionl here: the unit sphere is not necessrily compct in infinite dimensionl spces