MATH 4733, HOMEWORK 6, SOLUTIONS (DUE WEDNESDAYOCTOBER 14)

T. PRZEBINDA

1. Problem 2(a), page 300 in the book.

You integrate as explained in class and get

fX ∗ fY (x) =

14x+ 1

2for − 2 ≤ x ≤ 0,

−14x+ 1

2for 0 ≤ x ≤ 2,

0 otherwise.

2. Problem 5(a), page 302 in the book.

A straightforward integration shows that

fX ∗ fY (z) =λµ

µ− λ(e−λz − e−µz

). (1)

3. Fix a constant k > 0 and let

ft(x) =1√

4πkte−

x2

4kt (t > 0, x ∈ R)

(This is a normal density with the mean zero and the variance equal 2k.) For a differen-tiable, bounded integrable function g(x) set u(t, x) = g ∗ ft(x). Check that u satisfies theheat equation

∂tu(t, x) = k∂2xu(t, x)

Done in class (for k = 12.)

4. Keep the notation of Problem 3. Fix ε > 0. Show that

limt→0

max|x|≥ε|ft(x)| = 0.

Since the exponential is a decreasing function and since ft is positive,

max|x|≥ε|ft(x)| = ft(ε).

1

2 T. PRZEBINDA

By l’Hospital’s rule

limt→0

t−12

eε2

t

= limt→0

−12t−

12

− ε2

t2eε2

t

= limt→0

12t32

ε2eε2

t

= 0.

5. Deduce from the previous problem that

limt→0

∫|x|>ε

ft(x)g(z − x) dx = 0

This follows from Problem 4 because∣∣∣∣∫|x|>ε

ft(x)g(z − x) dx

∣∣∣∣ ≤ ∫|x|>ε|ft(x)g(z − x)| dx ≤ max

|x|≥ε|ft(x)|

∫R|g(x)| dx

6. Check that ∫Rft(x)g(z − x) dx− g(z) =

∫Rft(x)(g(z − x)− g(z)) dx.

This is obvious because ∫Rft(x) dx = 1.

7. Check that for any ε > 0,∣∣∣∣∫Rft(x)g(z − x) dx− g(z)

∣∣∣∣ ≤ max|x|≤ε|g(z − x)− g(z)|+

∣∣∣∣∫|x|>ε

(ft(x)g(z − x)− g(z)) dx

∣∣∣∣ .We see from Problem 6 that∫

Rft(x)g(z − x) dx− g(z)

=

∫|x|≤ε

ft(x)(g(z − x)− g(z)) dx+

∫|x|>ε

ft(x)(g(z − x)− g(z)) dx

Also, ∣∣∣∣∫|x|≤ε

ft(x)(g(z − x)− g(z)) dx

∣∣∣∣ ≤ ∫|x|≤ε

ft(x) max|x|≤ε|g(z − x)− g(z)| dx

≤∫Rft(x) dxmax

|x|≤ε|g(z − x)− g(z)| = max

|x|≤ε|g(z − x)− g(z)|

MATH 4733, HOMEWORK 6, SOLUTIONS (DUE WEDNESDAY OCTOBER 14) 3

and we are done.

8. Deduce from Problem 7 that

limt→0

u(t, z) = g(z).

We need to show that for any ε1 > 0 there is δ > 0 such that

|u(t, z)− g(z)| < ε1 if t < δ. (2)

Since g is continuous we may choose ε so that

max|x|≤ε|g(z − x)− g(z)| < ε1

2.

We know from Chebyshev that ∫|x|>ε

ft(x) dx ≤ 2kt

ε2.

Hence, there is δ > 0 such that∣∣∣∣∫|x|>ε

ft(x)(g(z − x)− g(z)) dx

∣∣∣∣ ≤ 2kt

ε22 maxx∈R|g(x)| < ε1

2if t < δ.

By adding the last two inequalities we get (2).