Question

In: Advanced Math

(a) Let M be a Cr submanifold of Rn, and let f : M → R...

(a) Let M be a Cr submanifold of Rn, and let f : M → R be a Cr function. Show
there is an open neighbourhood V of M in Rn and a Cr function g : V → R such
that f = g|M.
(b) Show that, if M is a closed subset of Rn, then we can take V = Rn
.
(c) Can we take V = Rn in general? Why or why not?
We've just learned partition of unity ,please don't use any high-level therom!!

Solutions

Expert Solution

Answer:-

Consider a subset M of RN is a k-dimensional submanifold if for every point x in M there exists a neighbourhood V of x in RN , an open set U ⊆ Rk, and a smooth map ξ : U → RN such that ξ is a homeomorphism onto M ∩ V , and Dyξ is injective for every y ∈ U.

  


Related Solutions

Let f : Rn → R be a differentiable function. Suppose that a point x∗ is...
Let f : Rn → R be a differentiable function. Suppose that a point x∗ is a local minimum of f along every line passes through x∗; that is, the function g(α) = f(x^∗ + αd) is minimized at α = 0 for all d ∈ R^n. (i) Show that ∇f(x∗) = 0. (ii) Show by example that x^∗ neen not be a local minimum of f. Hint: Consider the function of two variables f(y, z) = (z − py^2)(z...
Let A ⊆ R, let f : A → R be a function, and let c...
Let A ⊆ R, let f : A → R be a function, and let c be a limit point of A. Suppose that a student copied down the following definition of the limit of f at c: “we say that limx→c f(x) = L provided that, for all ε > 0, there exists a δ ≥ 0 such that if 0 < |x − c| < δ and x ∈ A, then |f(x) − L| < ε”. What was...
Let R be a ring and f : M −→ N a morphism of left R-modules. Show that:
Let R be a ring and f : M −→ N a morphism of left R-modules. Show that: c) K := {m ∈ M | f(m) = 0} satisfies the Universal Property of Kernels. d) N/f(M) satisfies the Universal Property of Cokernels. Q2. Show that ZQ :a) contains no minimal Z-submodule
Let R be a ring and f : M −→ N a morphism of left R-modules. Show that:
Let R be a ring and f : M −→ N a morphism of left R-modules. Show that: c) K := {m ∈ M | f(m) = 0} satisfies the Universal Property of Kernels. d) N/f(M) satisfies the Universal Property of Cokernels.  
Let R be a UFD and let F be a field of fractions for R. If...
Let R be a UFD and let F be a field of fractions for R. If f(α) = 0, where f ∈ R [x] is monic and α ∈ F, show that α ∈ R NOTE: A corollary is the fact that m ∈ Z and m is not an nth power in Z, then n√m is irrational.
Let f : R → R be a function. (a) Prove that f is continuous on...
Let f : R → R be a function. (a) Prove that f is continuous on R if and only if, for every open set U ⊆ R, the preimage f −1 (U) = {x ∈ R : f(x) ∈ U} is open. (b) Use part (a) to prove that if f is continuous on R, its zero set Z(f) = {x ∈ R : f(x) = 0} is closed.
Let A ∈ Rn×n be symmetric and positive definite and let C ∈ Rn×m. Show:, rank(C^TAC)...
Let A ∈ Rn×n be symmetric and positive definite and let C ∈ Rn×m. Show:, rank(C^TAC) = rank(C),
2 Let F be a field and let R = F[x, y] be the ring of...
2 Let F be a field and let R = F[x, y] be the ring of polynomials in two variables with coefficients in F. (a) Prove that ev(0,0) : F[x, y] → F p(x, y) → p(0, 0) is a surjective ring homomorphism. (b) Prove that ker ev(0,0) is equal to the ideal (x, y) = {xr(x, y) + ys(x, y) | r,s ∈ F[x, y]} (c) Use the first isomorphism theorem to prove that (x, y) ⊆ F[x, y]...
6. (a) let f : R → R be a function defined by f(x) = x...
6. (a) let f : R → R be a function defined by f(x) = x + 4 if x ≤ 1 ax + b if 1 < x ≤ 3 3x x 8 if x > 3 Find the values of a and b that makes f(x) continuous on R. [10 marks] (b) Find the derivative of f(x) = tann 1 1 ∞x 1 + x . [15 marks] (c) Find f 0 (x) using logarithmic differentiation, where f(x)...
5. (a) Let f : R \ {−1} → R, f(x) = x+1. Show that f...
5. (a) Let f : R \ {−1} → R, f(x) = x+1. Show that f is injective, but not surjective. (b) Suppose g : R\{−1} → R\{a} is a function such that g(x) = x−1, where a ∈ R. Determine x+1 a, show that g is bijective and determine its inverse function.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT