Math 509: Advanced Analysis

Homework 2

Lawrence Tyler Rush

<me@tylerlogic.com>

February 1, 2015

http://coursework.tylerlogic.com/courses/upenn/math509/homework02

1 Provide a complete proof of the Chain Rule

Let f : R^m → Rⁿ and g : Rⁿ → R^p be differentiable functions. Furthermore, let x₀ ∈ Rⁿ, y₀ ∈ R^m, and z₀ ∈ R^p such that f(x₀) = y₀ and g(y₀) = z₀. For ease of notation, we asssume that, without loss of generality, the points x₀ ∈ R^m, y₀ ∈ Rⁿ and z₀ ∈ R^p are the origins, respectively, of R^m, Rⁿ, and R^p. We may safely assume this because this situation can be obtained with a simple translation of the functions f and g, which leaves there differentiability, namely the limit formula that makes them differentiable, unaffected.

Now define the linear maps L = f′(x₀) and M = g′(y₀) to be the derivatives of f and g, respectively, at x₀ and y₀. Thus we have

f-(x)-- L-(x) f(x0-+x-)--f(x0)---L(x) |x| = |x| → 0 as x → 0

and

g(y)--M-(y)= g(y0-+y)---g(y0)--M-(y)→ 0 as y → 0 |y| |y|

via the combination of the differentiability of f and g and our assumption that x₀, f(x₀) = y₀, and g(y₀) = z₀ are all at the origin of their respective spaces. These two equations then yield the following sequence of equations for sufficiently small x ∈ R^m

|gf (x) - M L(x)| = |gf(x)- (M f(x )- M f(x)) - M L(x)| = |gf(x)- M f(x)+ M f(x)- M L(x)| ≤ |g- M ||f(x)|+ |M ||f (x) - L (x )| < ε|f(x)|+ |M |(ε|x|)

Dividing the last line by |x| then yields

|gf-(x)-- M-L(x)| |f-(x)| |x| < ε |x| + ε|M |

Hence if we can show that |f(x|x)|| is bounded as |x|→ 0, then the previous inequality implies that |gf(x)-|xM|L(x)| can be arbitrarily bounded as |x|→ 0. Such a property on |gf(x)-ML-(x)|
|x| proves our theorem since

|gf (x0 + x)- gf(x0) - M L(x)| |gf(x) - M L(x)| ------------|x|-------------= ------|x-|-----

as we assumed x₀ is the origin.

So to, finally, prove the boundedness of |f(x)|∕|x| we see that

|f(x)| |f(x)+ (L(x)- L(x))| |L(x)|+|f(x)- L(x)| |L(x)| |f(x)- L(x)| -----= -------------------≤ -------------------= ----- + ----------- |x| |x| |x| |x| |x|

but |L(x)|∕|x| < |L| and |f(x)-L(x)|
|x| → 0 as |x|→ 0. Hence |f(x)|
|x| is bounded as |x|→ 0 as desired.

2 Provide a complete proof of the equality mixed partials (for m = 2)

Let U be an open set in R² and f : U → R a function whose partial derivatives of orders one and two exist and are continuous on U. So for arbitrary h ∈ R, define the function g_h : U → R by

gh(x,y) = f (x + h,y)- f(x,y)

and then define E : U → R by

E (h,k) = gh(x,y +k) - gh(x,y)

Due to the definition of g_h we can apply the mean value theorem (MVT) to E twice, but in two different ways:

We can apply the MVT to g_h in E to get $∂ E(h,k) = k∂y-gh(x,y + t1k)$
for some 0 ≤ t₁ ≤ 1. We next expand g_h in the result to obtain
$E(h,k) = k ∂-(f(x+ h,y + t1k)- f(x,y +t1k)) ∂y$
and finally apply the MVT once more now to f to get
$-∂2-- E(h,k) = hk ∂y∂xf(x +s1h,y +t1k)$ (2.1)

for some 0 ≤ s₁ ≤ 1.
Alternatively, we can first expand both g_h components of E obtaining $E (h,k) = (f(x + h,y+ k)- f(x,y+ k))- (f(x+ h,y)- f(x,y))$
and then apply the MVT first to the resulting expansion of the first component and then a second time to the resulting expansion of the second component to get
$E(h,k) = (f(x+ h,y+ k)- f (x,y + k))- (f(x +h,y) - f (x,y)) ∂ ∂ = h∂x-f(x+ s2h,y+ k)- h∂x-f(x+ s2h,y) ∂ = h∂x-(f(x+ s2h,y+ k)- f(x + s2h,y))$
for some 0 ≤ s₂ ≤ 1. We then apply the MVT once more yielding
$( ) 2 E (h,k ) = h-∂ k ∂-f(x+ s2h,y+ t2k) = hk-∂--f (x + s2h,y + t2k) ∂x ∂y ∂x∂y$ (2.2)

for some 0 ≤ t₂ ≤ 1.

Therefore equations 2.1 and 2.2 yield

∂2 E (h,k) ∂2 ∂y-∂xf(x+ s1h,y+ t1k) =--hk-- = ∂x-∂yf(x+ s2h,y+ t2k)

(2.3)

Now in light of the definition of E, we see that E is also continuous in both h and k so that letting h → 0 followed by letting k → 0 will have the same value as letting k → 0 followed by letting h → 0. Thus we can let h → 0 and k → 0, safely in either order, in equation 2.3 to obtain