Topics In Analysis

5.10 Proofs Of Theorems

This section contains the proofs of the theorems which were just stated without proof.

Proof: Begin with 1.) Let ε > 0 be given. By assumption, there exist δ₁ > 0 such that whenever

< δ₁, it follows < and there exists δ₂ > 0 such that whenever < δ₂, it follows that < . Then let 0 < δ ≤ min. If < δ, then everything happens at once. Therefore, using the triangle inequality

|af (x)+ bf (x)− (ag(y)+ bg (y ))|

	≤ |a| |f (x)− f (y)| + |b| |g(x)− g (y )|
	< |a| ( ) -----ε------- 2(|a|+ |b|+ 1) + |b| ( ) ------ε------ 2 (|a|+ |b|+ 1) < ε.

Now begin on 2.) There exists δ₁ > 0 such that if

< δ₁, then

|f (x)− f (y)| < 1.

Therefore, for such y,

|f (y )| < 1 + |f (x )|.

It follows that for such y,

|fg(x)− fg(y)| ≤ |f (x)g(x)− g (x )f (y)|+ |g(x)f (y)− f (y)g (y)|

	≤ |g (x)| |f (x)− f (y )| + |f (y)| |g(x)− g(y)|
	≤ (1 + |g (x)|+ |f (y)|) [|g(x)− g(y)|+ |f (x )− f (y)|] .

Now let ε > 0 be given. There exists δ₂ such that if

< δ₂, then

|g(x)− g(y)| <---------ε---------, 2 (1 + |g (x)|+ |f (y)|)

and there exists δ₃ such that if

< δ₃, then

ε |f (x )− f (y)| < 2(1+-|g(x)|+-|f-(y-)|)

Now let 0 < δ ≤ min

. Then if < δ, all the above hold at once and

|fg(x)− fg (y)| ≤

	(1+ |g(x)|+ |f (y)|) [|g(x) − g (y )|+ |f (x)− f (y )|]
	< (1 + |g (x)|+ |f (y)|) ( ) ---------ε---------+ --------ε---------- 2 (1 + |g(x)|+ |f (y)|) 2(1+ |g (x )|+ |f (y)|) = ε.

This proves the first part of 2.) To obtain the second part, let δ₁ be as described above and let δ₀ > 0 be such that for

< δ₀,

|g (x) − g (y)| < |g(x)|∕2

and so by the triangle inequality,

− |g(x)|∕2 ≤ |g (y )|− |g(x)| ≤ |g (x )|∕2

which implies

≥∕2, and < 3∕2.

Then if

< min,

|| || ||f-(x-)− f-(y-)|| g (x ) g (y)	= || || ||f (x)g(y)-− f-(y)g(x)|| g(x)g (y )
	≤ |f-(x-)g((y)−-f ()y)g-(x-)| |g(x)|2- 2
	= 2|f (x)g(y)−-f-(y-)g(x)| |g(x)|2

	≤ --2--- |g(x)|2 [|f (x)g(y)− f (y )g(y)+ f (y)g (y )− f (y)g(x)|]
	≤ 2 |g(x)|2 [|g (y )||f (x)− f (y )|+ |f (y)||g (y) − g(x)|]
	≤ --2--2 |g(x)| [ ] 3|g(x)||f (x) − f (y)|+ (1 + |f (x)|)|g (y )− g(x)| 2
	≤ --2--- |g(x)|2 (1 +2 |f (x )|+ 2|g(x)|) [|f (x)− f (y )|+ |g(y)− g(x)|]
	≡ M [|f (x )− f (y)|+ |g(y) − g (x )|]

where

M ≡ ---2-- (1 + 2|f (x)|+ 2|g (x )|) |g(x)|2

Now let δ₂ be such that if

< δ₂, then

ε |f (x)− f (y)| <-M −1 2

and let δ₃ be such that if

< δ₃, then

|g(y)− g(x)| < εM −1. 2

Then if 0 < δ ≤ min

, and < δ, everything holds and

||f (x ) f (y )|| ||g-(x-) − g-(y)-|| ≤ M [|f (x) − f (y)|+ |g(y)− g (x )|]

[ ] < M εM −1 + εM −1 = ε. 2 2

This completes the proof of the second part of 2.) Note that in these proofs no effort is made to find some sort of “best” δ. The problem is one which has a yes or a no answer. Either it is or it is not continuous.

Now begin on 3.). If f is continuous at x, f

∈ D ⊆ F^p, and g is continuous at f,then g ∘ f is continuous at x. Let ε > 0 be given. Then there exists η > 0 such that if < η and y ∈ D, it follows that < ε. It follows from continuity of f at x that there exists δ > 0 such that if < δ and z ∈ D, then < η. Then if < δ and z ∈ D ⊆ D, all the above hold and so

|g (f (z))− g (f (x ))| < ε.

This proves part 3.)

Part 4.) says: If f =

: D → F^q, then f is continuous if and only if each f_k is a continuous F valued function. Then

|fk(x)− fk (y)| ≤ |f (x)− f (y)|

( ) ∑q 2 1∕2 ≡ |fi(x)− fi(y)| i=1

∑q ≤ |fi(x) − fi(y)|. (5.10.11) i=1

(5.10.11)

Suppose first that f is continuous at x. Then there exists δ > 0 such that if

< δ, then < ε. The first part of the above inequality then shows that for each k = 1,,q, < ε. This shows the only if part. Now suppose each function, f_k is continuous. Then if ε > 0 is given, there exists δ_k > 0 such that whenever < δ_k

|fk(x) − fk (y )| < ε∕q.

Now let 0 < δ ≤ min

. For < δ, the above inequality holds for all k and so the last part of 5.10.11 implies

|f (x)− f (y)|	≤∑ i=1q |fi(x )− fi(y )|
	< ∑ i=1q ε q = ε.

This proves part 4.)

To verify part 5.), let ε > 0 be given and let δ = ε. Then if

< δ, the triangle inequality implies

|f (x)− f (y )|	= ||x|− |y||
	≤ |x − y| < δ = ε.

This proves part 5.) and completes the proof of the theorem.

Here is a multidimensional version of the nested interval lemma.

The following definition is similar to that given earlier. It defines what is meant by a sequentially compact set in F^p.

It turns out the sequentially compact sets in F^pare exactly those which are closed and bounded. Only half of this result will be needed in this book and this is proved next. First note that ℂ can be considered as ℝ². Therefore, ℂ^p may be considered as ℝ^2p.

Proof: Let

⊆ C. Then let a_n = . It follows the real and imaginary parts of the terms of the sequence, _n=1^∞ are each contained in some sufficiently large closed bounded interval. By Theorem 2.0.3 on Page 43, there is a subsequence of the sequence of real parts of _n=1^∞ which converges. Also there is a further subsequence of the imaginary parts of _n=1^∞ which converges. Thus there is a subsequence, n_k with the property that a_j^n_k converges to a point, a _j ∈ F. Taking further subsequences, one obtains the existence of a subsequence, still called n_k such that for each r = 1,,p, a_r^n_k converges to a point, a _r ∈ F as k →∞. Therefore, letting a ≡,lim_k→∞a^n_k = a. Since C is closed, it follows a ∈ C. This proves the theorem.

Here is a proof of the extreme value theorem.

Proof: Let M = sup

. Recall this means +∞ if f is not bounded above and it equals the least upper bound of these values of f if f is bounded above. Then there exists a sequence, such that f → M. Since C is sequentially compact, there exists a subsequence, x_nk, and a point, x ∈ C such that x_nk → x. But then since f is continuous at x, it follows from Theorem 5.7.10 on Page 306 that f = lim_k→∞f = M. This proves f achieves its maximum and also shows its maximum is less than ∞. Let x₂ = x. The case of a minimum is handled similarly.

Recall that a function is uniformly continuous if the following definition holds.

Proof: If this is not so, there exists ε > 0 and pairs of points, x_n and y_n satisfying

< 1∕n but ≥ ε. Since C is sequentially compact, there exists x ∈ C and a subsequence, satisfying x_nk → x. But < 1∕k and so y_nk → x also. Therefore, from Theorem 5.7.10 on Page 306,

ε ≤ lim |f (xnk) − f (ynk )| = |f (x)− f (x)| = 0, k→∞

a contradiction. This proves the theorem.