Homework 11 <me@tylerlogic.com>
December 4, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework11

### 1

Let L : X Y be linear map between vector spaces X and Y such that x1,x2 X are solutions, respectively, to

for some y1,y2 Y . Furthermore, let z0 be a solution to Lx = 0.

#### (a) Find a solution for Lx = 3y1

The vector 3x1 is a solution since L(3x1) = 3(Lx1) = 3y1.

#### (b) Find a solution for Lx = -5y2

The vector -5x2 is a solution since L(-5x2) = -5(Lx2) = -5y2.

#### (c) Find a solution for Lx = 3y1- 5y2

The vector 3x1 - 5x2 is a solution since L(3x1 - 5x2) = 3(Lx1) - 5(Lx2) = 3y1 - 5y2.

#### (d) Find a solution other than z and 0 for Lx = 0

The vector 2z is a solution since L(2z) = 2(Lz) = 2(0) = 0.

#### (e) Find two solutions of Lx = y1

Both x1 + z and x1 + 2z are solutions since L(x1 + z) = Lx1 + Lz = y1 + 0 = y1 and L(x1 + 2z) = Lx1 + 2Lz = y1 + 2(0) = y1.

#### (f) Find another solution to Lx = 3y1- 5y2

The vector 3x1 - 5x2 + z is another solution since L(3x1 - 5x2 + z) = 3(Lx1) - 5(Lx2) + Lz = 3y1 - 5y2 + 0 = 3y1 - 5y2.

### 5

Since K(x,y) is continuous on [0,1] × [0,1] and [0,1] × [0,1] is compact in 2, then K is uniformly continuous on [0,1] × [0,1]. Let ε > 0. Then for any x,z there exists a δ > 0 such that |x-z| < δ implies |K(x,y) - K(z,y)| < ε for all y . This implies that

so that h is continuous.

### 6

If we rewrite

as αex where we define the constant α = 01e-yu(y)dy we can simplify u(x) as

 (6.1)

From this simplification we have e-xu(x) = e-xf(x) + λα which, after integrating both sides with respect to x over [0,1], implies

Combining the last line of the above equation with equation 6.1 we end up with

Hence we have a unique solution for

so long as λ0.

### 7

Not worded properly.