Analysis

10.1 Mollifiers And Partitions Of Unity

From Theorem 9.2.3, every function in L¹

can be approximated by one in C_c but even more incredible things can be said. In fact, you can approximate an arbitrary function in L¹ with one which is infinitely differentiable having compact support. From now on, the notation will refer to the Euclidean norm,

(∑n )1∕2 |x | = |xk|2 . k=1

This is very important in partial differential equations. I am just giving a short introduction to this concept here. Consider the following example.

You show this by verifying the partial derivatives all exist and are continuous. The only place this is hard is when

= r. It is left as an exercise. You might consider a simpler example,

{ e− 1∕x2 if x ⁄= 0 f (x) = 0 if x = 0

and reduce the above to a consideration of something like this simpler case.

Proof: Pick z ∈ U and let r be small enough that B

⊆ U. Then let ψ ∈ C_c^∞ ⊆ C_c^∞ be the function of the above example. ■

∫ f(x,y)dm_p(y) will mean x is fixed and the function y → f(x,y) is being integrated. To make the notation more familiar, dx is written instead of dm_p(x).

The following is an important property of the convolution.

Proof: This follows right away from the change of variables formula. In the left, let x − y ≡ u. Then the left side equals

∫ f (x − u)g(u)du

because the absolute value of the determinant of the derivative is 1. Now replace u with y and this proves the proposition. ■

The following lemma will be useful in what follows. It says among other things that one of these very unregular functions in L_loc¹

is smoothed out by convolving with a mollifier.

Proof: Consider the difference quotient for calculating a partial derivative of f ∗ g.

∫ f-∗g-(x-+-tej)-− f-∗g-(x-)= f(y)g(x-+-tej −-y-)−-g(x−-y)dy. t t

Using the fact that g ∈ C_c^∞

, the quotient,

g(x-+-tej −-y)−-g(x-− y-), t

is uniformly bounded. To see this easily, use Theorem 5.5.2 on Page 279 to get the existence of a constant, M  depending on

max {||Dg (x)|| : x ∈ ℝp}

such that

|g(x + tej − y)− g(x− y )| ≤ M |t|

for any choice of x and y. Therefore, there exists a dominating function for the integrand of the above integral which is of the form C

X_K where K is a compact set depending on the support of g. It follows from the dominated convergence theorem the limit of the difference quotient above passes inside the integral as t → 0 and so

∂ ∫ ∂ ---(f ∗g)(x) = f(y )---g(x− y )dy. ∂xj ∂xj

Now letting

g play the role of g in the above argument, a repeat of the above reasoning shows partial derivatives of all orders exist. A similar use of the dominated convergence theorem shows all these partial derivatives are also continuous.

It remains to verify the claim about the mollifier. Let x ∈ U and let m be large enough that B

⊆ U. Then

∫ |f ∗ g(x)− f (x)| ≤ |f (x − y )− f (x)|ψm (y )dy B(0,1m)

By continuity of f at x, for all m sufficiently large, the above is dominated by

∫ ε ψm (y)dy = ε B (0,1m)

and this proves the claim.

Now consider the formula in the case where f ∈ C¹

. Using Proposition 10.1.6,

f ∗ψm (x + hei)− f ∗ψm (x) ------------h------------ =

1 (∫ ∫ ) -- 1 f (x+hei− y)ψm (y)dy− 1 f (x − y)ψm (y)dy h B (0,m) B(0,m)

∫ = (f (x+hei-− y)−-f-(x−-y))ψ (y)dy B(0,1m) h m

Now letting m be small enough and using the continuity of the partial derivatives, it follows the difference quotients are uniformly bounded for all h sufficiently small and so one can apply the dominated convergence theorem and pass to the limit obtaining

∫ ℝp fxi (x − y)ψm (y)dy ≡ fxi ∗ ψm (x )

Finally, consider the last claim.

|∫ | ∫ || f (x− y)ψ (y)dy − f (x)||≤ |f (x − y)− f (x)|ψ (y)dy | U m | U m

By uniform continuity of f for m large enough,

< ε for all x whenever < 1∕m and so for all m large enough, the above integral is dominated by

∫ ∥f ∗ψm − f ∥ ≤ εψm (y)dy = ε. ■ B(0,1∕m)

Proof: Let r > 0 be small enough that K + B(0,3r) ⊆ U. The symbol, K + B(0,3r) means

{k + x : k ∈ K and x ∈ B(0,3r)}.

Thus this is simply a way to write

∪{B (k,3r) : k ∈ K} .

Think of it as fattening up the set, K. Let K_r = K + B(0,r). A picture of what is happening follows. Consider X_Kr ∗ ψ_m where ψ_m is a mollifier. Let m be so large

that < r. Then from the definition of what is meant by a convolution, and using that ψ_m has support in B, X_Kr ∗ψ_m = 1 on K and its support is in K + B, a bounded set. Now using Lemma 10.1.7, X_Kr ∗ ψ_m is also infinitely differentiable. Therefore, let h = X_Kr ∗ ψ_m.

As to the existence of the open set W, let it equal the closed and bounded set h⁻¹

. This proves the theorem. ■

The following is the remarkable theorem mentioned above. First, here is some notation.

Proof: Let f ∈ L¹(ℝ^p) and let ε > 0 be given. Choose g ∈ C_c(ℝ^p) such that

∫ |g − f|dmp < ε∕2

This can be done by using Theorem 9.2.3. Now let

∫ ∫ gm(x) = g∗ψm (x) ≡ g(x− y) ψm (y) dmp(y) = g(y)ψm (x− y)dmp (y)

where {ψ_m} is a mollifier. It follows from Lemma 10.1.7 g_m ∈ C_c^∞(ℝ^p). It vanishes if x

spt(g) + B(0,).

whenever m is large enough. This follows because since g has compact support, it is uniformly continuous on ℝ^p and so if η > 0 is given, then whenever is sufficiently small,

|g (x) − g(x− y)| < η

for all x. Thus, since g has compact support, if y is small enough, it follows

∫ |g − gy|dmp (x ) < ε∕2.

There is no measurability problem in the use of Fubini’s theorem because the function

(x,y) → |g(x) − g(x − y)|ψm (y)

is continuous. Thus when m is large enough,

∫ ∫ ∫ ε ε |f − gm|dmp ≤ |f − g|dmp + |g− gm|dmp < 2 + 2 = ε.

This proves the theorem. ■ 

Another important application of Theorem 10.1.8 has to do with a partition of unity.

The thing about locally finite collection of sets is that the closure of their union equals the union of their closures. This is clearly true of a finite collection.

Proof: It is obvious ⊇ holds in the above claim. It remains to go the other way. Suppose then that p is a limit point of ∪ℋ and p

∪ℋ. There exists r > 0 such that B has nonempty intersection with only finitely many sets of ℋ say these are H₁,,H_m. Then I claim p must be a limit point of one of these. If this is not so, there would exist r^′ such that 0 < r^′ < r with B having empty intersection with each of these H_i. But then p would fail to be a limit point of ∪ℋ. Therefore, p is contained in the right side. It is clear ∪ℋ is contained in the right side and so This proves the lemma. ■

A good example to consider is the rational numbers each being a set in ℝ. This is not a locally finite collection of sets and you note that ℚ = ℝ≠∪

. By contrast, ℤ is a locally finite collection of sets, the sets consisting of individual integers. The closure of ℤ is equal to ℤ because ℤ has no limit points so it contains them all.

A version of the following lemma is valid for locally finite coverings, but we are only using it when the covering is finite.

Proof: Let K₁ = K ∖∪_i=2ⁿV _i. Thus K₁ is compact because K₁ ⊆ V ₁. Let W₁ be an open set having compact closure which satisfies

--- K1 ⊆ W1 ⊆ W 1 ⊆ V1

Thus W₁,V ₂,

,V _n covers K and W₁ ⊆ V ₁. Suppose W₁,,W_r have been defined such that W_i ⊆ V _i for each i, and W₁,,W_r,V _r+1,,V _n covers K. Then let

( n ) ( r ) Kr+1 ≡ K ∖( ∪i=r+2Vi ∪ ∪ j=1Wj ).

It follows K_r+1 is compact because K_r+1 ⊆ V _r+1. Let W_r+1 satisfy

--- --- Kr+1 ⊆ Wr+1 ⊆ W r+1 ⊆ Vr+1,W r+1 is compact

Continuing this way defines a sequence of open sets

_i=1ⁿ having compact closures with the property

--- Wi ⊆ Vi,K ⊆ ∪ni=1Wi.

Note

_i=1ⁿ is locally finite because the original list, _i=1ⁿ was locally finite. Now let U_i be open sets which satisfy

--- -- -- W i ⊆ Ui ⊆ U i ⊆ Vi,U i is compact.

Similarly,

_i=1ⁿ is locally finite.

Now ∪_i=1ⁿW_i = ∪_i=1ⁿW_i and so it is possible from Theorem 10.1.8 to define ϕ_i and γ, infinitely differentiable functions having compact support such that

-- n --- n U i ≺ ϕi ≺ Vi, ∪i=1W i ≺ γ ≺ ∪ i=1Ui.

Now define

{ ∑n ∑n ψi(x) = γ(x∑)nϕi(x)∕ j=1ϕj(x)if j=1ϕj(x) ⁄= 0, 0if j=1ϕj(x) = 0.

If x is such that ∑ _j=1ⁿϕ_j(x) = 0, then x

∪_i=1ⁿU_i because ϕ_i equals one on U_i. Consequently γ = 0 for all y near x thanks to the fact that ∪_i=1ⁿU_i is closed and so ψ_i(y) = 0 for all y near x. Hence ψ_i is infinitely differentiable at such x. If ∑ _j=1ⁿϕ_j(x