Math 509: Advanced Analysis

Homework 1

Lawrence Tyler Rush

<me@tylerlogic.com>

January 24, 2015

http://coursework.tylerlogic.com/courses/upenn/math509/homework01

1 Lecture Notes Problem 1

Define f : ℝ² → ℝ by

{ 0 if (x,y) = (0,0) f(x,y) = x2x+yy2- otherwise

This then yields the following partial derivatives f_x : ℝ² → ℝ and f_y : ℝ² → ℝ for nonzero points

f (x,y) =---y---- --2x2y--- x x2 +y2 (x2 + y2)2

and

2 fy(x,y) =-2-x-2-- --22xy-2-2 x +y (x + y )

Thus there are no points in ℝ² - (0,0) for which f_x and f_y don’t exist. It remains to be proven that the partial derivatives exist at the origin. Since the following expression evaluates to zero:

f(h,0)- f(0,0) 0 fx = lhim→0------h-------= lhim→0h-= hli→m0 0 = 0

then the partial derivative f_x exists at (0,0). A symmetrical argument proves the existence of f_y at (0,0).

2 Lecture Notes Problem 2

Let U be a convex set in ℝ^m and f : U → ℝⁿ a differentiable map such that there exists a real M with |f′(x)|≤ M for all x ∈ U. Let p,q ∈ U and define two maps ϕ : (0,1) → U, g : (0,1) → ℝⁿ where (0,1) ⊂ ℝ is an open interval, by

ϕ(t) = tp +(1 - t)q and g(t) = f (ϕ(t))

Note that g is well defined because the range of ϕ is a subset of U due to U being convex. So taking the derivative of g we get g′(t) = f′(ϕ(t))ϕ′(t) by the chain rule so that g′(t) = f′(ϕ(t))(p - q) due to the fact that ϕ the function for the points on the line segment connecting p and q. Taking the norm of both sides then yields

′ ′ |g (t)| = |f (ϕ(t))||p- q| ≤ M |p- q|

(2.1)

Now Rudin Theorem 5.19 tells us that |g(1) - g(0)|≤ (1 - 0)|g′(t)| for all t ∈ [0,1], which implies

|f(p) - f (q)| = |f(ϕ (1))- f(ϕ(0))| = |g(1)- g(0)| ≤ |g′(t)|

(2.2)

Combining equations 2.1 and 2.2 gives to us the desired result:

|f(p)- f(q)| ≤ M |p- q|

3 Lecture Notes Problem 4

Let f : ℝ² → ℝ be defined by

2 2 f(x,y) = 2y - x(x - 1)

Then the partial derivatives of f are

fx = - (x- 1)2 - 2x(x- 1) = f = 4y y

so that f_x is zero at x = 1
3

,1 and f_y is zero when y = 0. Thus the critical points are ( 1
3

,0) and (1,0). Finally, because the Hessian matrix is

( fxx fxy) ( 4- 6x 0 ) fyx fyy = 0 4

then

( ) 1 2 0 det(H (3,0)) = det 0 4 = 8

and

( ) det(H(1,0)) = det - 2 0 = - 8 0 4

implying that the point ( 1
3 ,0) is a local minimum and (1,0) is a saddle point.

We see this function graphed with the following sketch:

4 Lecture Notes Problem 5

Let f : ℝ² → ℝ be defined by

2 2 f(x,y) = ax + 2bxy + cy

Then the partial derivatives of f are

fx = 2ax+ 2by f = 2bx+ 2cy y

Thus we get that f_x = 0 whenever ax = -by and f_y = 0 whenever bx = -cy. Hence the only critical point is (0,0). The Hessian matrix

( fxx fxy ) ( 2a 2b ) fyx fyy = 2b 2c

is completely independent of x and y. So because det(H) = 4ac- 4b² then the critical point (0,0) will be a local minimum if ac > b² and a,c > 0, a local maximum if ac > b² and a,c < 0, saddle point if ac < b², nondegenerate if ac - b²≠0, and degenerate if ac = b².