LECTURE 6, 18.155, 27 SEPTEMBER 2011
(1) Returning to the Schwartz’ space S(Rn ) recall that we defined
the space of tempered distributions by
S 0 (Rn ) = {u : S(Rn ) −→ C; u is linear and continuous.}.
Then we embedded S(Rn ) into S 0 (Rn ) by identifying φ ∈ S(Rn )
with the distribution, denoted temporarily Uφ ,
Z
Uφ (ψ) =
φ(x)ψ(x).
Rn
Continuity is an estimate we discussed earlier
Z
2 n
|Uφ (ψ)| ≤ sup |φ(x)|·sup(1+|x| ) |ψ(x)|· (1+|x|2 )−n ≤ Ckφk∞ kψ|2n .
We know that this map is injective, since taking ψ = φ shows
that if Uφ = 0 in S 0 (Rn ), which means
R 2 precisely that Uφ (ψ) = 0
n
for all ψ ∈ S(R ) then Uφ (φ) = |φ| = 0 so φ = 0 a.e.
Exercise 1. Show that a similar argument extends to give an
injection of bounded continuous functions into S 0 (Rn ).
(2) Now, we want to extend this injection to all the spaces Lp (Rn );
in fact we are most interested in the cases p = 1 and 2 but the
general case is important (on the other hand I have not really
discussed L∞ (Rn ) so will exclude that for the moment, although
nothing too significant happens except that S(Rn ) is not dense
in L∞ (Rn ).
The basic fact which is clear from the estimate above is that
S(Rn ) ⊂ Lp (Rn ), 1 ≤ p < ∞ (and p∞.)
Indeed, the sequence φk (x) = φχ( xk ) where χ ∈ Cc∞ (Rn ), χ = 1
in |x| < 1 is absolutely summable in Lp and converges to φ ∈
S(Rn ). The same estimate as above holds
kφkLp ≤ Cp kφk2n , φ ∈ S(Rn )
in terms of our norms on S(Rn ).
(3) Think about the case p = 2. Then
(1)
L2 (Rn ) ,→ S 0 (Rn )
1
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LECTURE 6, 18.155, 27 SEPTEMBER 2011
where the embedding is given by
Z
f 7−→ Uf , Uf (φ) =
f (x)φ(x)dx, φ ∈ S(Rn )
Rn
and from what we have just seen and the Cauchy-Schwarz inequality
|Uf (φ)| ≤ kf kL2 kφkL2 ≤ Ckf kL2 kφk2n .
This gives the map in (1). Injectivity of this map follows
from the density of S(Rn ) in L2 (Rn ) since we can then choose
φl ∈ S(Rn ) such that φl → f in L2 (Rn ). So if Uf = 0 then
Uf (φ) = 0 for Rall φ ∈ S(Rn ) (which is what it means) and so
0 = Uf (φl ) → |f |2 shows that (1) is indeed injective.
(4) The same argument works for any 1 < p < ∞ just replacing
Cauchy-Schwarz by Hölder’s inequality. Thus if f ∈ Lp (Rn )
then f |f |p−2 (defined to be zero when f vanishes) is in Lq (Rn )
p
for the conjugate q = p−1
and the same argument can be used.
This proves injectivity, but I will not go through the details
since it also follows from the L1 case with only a little extra
thought.
(5) So, this argument does not work for p = 1 although it can be
made to do so. Still, I prefer to give a different, maybe more
involved, argument that leads us to convolution which we want
for plenty of other reasons.
(6) Time for a Lemma about integrable functions before we start
this.
Lemma 1 (Continuity in the mean). For any 1 ≤ p < ∞ and
f ∈ Lp (Rn ), translation gives a continuous map Rn 3 y 7−→
f (· − y) ∈ Lp (Rn ).
Proof. First check that f (x−y) is, as a function of x for y fixed,
in Lp (Rn ). This in fact follows from the translation-invariance of
Riemann and hence Lebesgue integrals. More precisely, if φk ∈
Cc0 (Rn ) is an approximating (absolutely summable) sequence for
f ∈ Lp (Rn ) then φk (· − y) is an approximating sequence in for
f (· − y).
So, the main claim here is the continuity, since both spaces
are metric space, this just means
Z
lim
|f (x) − f (x − y)|p dx = 0.
|y|→0
In fact, again because we are dealing with metricR topologies, it
suffices to show that for any sequence yp → 0, |f (x − yp ) −
LECTURE 6, 18.155, 27 SEPTEMBER 2011
3
f (x)|p → 0. Again, we can just use an approximating sequence,
φk as above. For a continuous function of compact support this
continuity in the mean in Lp is obvious, since φ(x − y) → φ(x)
uniformly as y → 0 (by uniform continuity). So, just break the
integral into three pieces using the triangle inquality
kfy − f kLp ≤ kfy − φk (· − y)kLp + kφk (· − y) − φk kLp + kφk − f kLp
where fy (x) = f (x − y). To make the LHS less than > 0 first
choose k so large that the first and last terms, which are in fact
equal by translation-invariantce, are less than /3 – this is the
approximation property. Then choose |y| small so the middle
term is less than + 3.
(7) So, the problem with the map L1 (Rn ) −→ S 0 (Rn ) is that we
have to think of some other way to show that we can ‘recover’
n
f ∈ L1 (R
R ) from the distribution Ufn. So, what we know is
Uf (φ) = f (x)φ(x) for every φ ∈ S(R ). This means in particular that we know the right side of
Z
(f ∗ φ)(y) = f (x)φ(y − x)dx
since φ(y − ·) ∈ S(Rn ) – since this is just a sign-reversal compared to what we did above. In fact we can see that f ∗ φ(y)
depends continuously on y (in fact it is infinitely differentiable).
This function is called the convolution of f and φ and we will
see it much more generally below.
Exercise 2. Show that ∂ α (f ∗ φ) ∈ L1 (Rn ) for all α.
We will get to this later but it follows from the discussion
below with a little thought.
(8) Recall that we showed the existence of χ ∈ Cc∞ (Rn ) which is
everywhere non-negative and has χ(x) = 1 near x = 0. It follows
that multiplying
0 ≤ µ ∈
R by a constant c > 0 there exists −n
∞
n
Cc (R ) with µ = 1. For such a µ set µ (x) = µ(x/),
> 0. The constant out the front ensures the properties
Z
∞
n
0 ≤ µ ∈ Cc (R ),
µ = 1, µ (x) = 0 in |x| > 2
which is all we really use in the argument below to prove that
f ∗ µ → f in L1 (Rn ) as ↓ 0.
First we need to show that f ∗ µ ∈ L1 (Rn ). In fact we circumvent this by assuming initially that f ∈ Cc0 (Rn ) and then
we use an approximation argument in general. For these special
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LECTURE 6, 18.155, 27 SEPTEMBER 2011
functions the integral defining the convolution clearly vanishes
for large y so f ∗ µ ∈ Cc0 and it is the convergence we want.
If ψk ∈ Cc0 (Rn ) then using the triangle inequality, changing
the order and variables of integration to x and z = y − x and
recalling the properties of µ
Z Z
Z Z
kψk ∗µ −ψl ∗µ kL1 =
(ψk (x)−ψl (x))µ (y−x)dx dy ≤
|ψk (x)−ψl (x)|µ (z)dxdz ≤ kψk −
Thus, if ψk is absolutely summable in L1 , so is ψk ∗ µ and it
converges pointwise to f ∗ µ if ψk is approximating for f. Thus
f ∗ µ ∈ L 1 .
That is, we need to consider the norm of the difference
Z
Z
kf − f ∗ µ kL1 =
f (y) − f (y)µ (y − x)dxdy.
The problem with this is that the two terms in side the integral
are so different. So we make the first look like the second by
using the integral property of µ :
Z Z
Z
kf − f ∗ µ kL1 =
f (y)µ (x − y)dx − f (x)µ (y − x)dxdy.
Here 1 has been replaced by the integral of µ after a change
of variable. Now, we can combine the two integrals into one to
get and use the triangle inequality
Z Z
Z Z
kf −f ∗µ kL1 =
(f (y)−f (x))µ (y−x)dx dy ≤
|f (y)−f (y−z)|µ (z)dydz
where I have also changed variable from x to z = y − x and
used the positivity of µ . In this integral, |z| ≤ 2 so we can use
continuity-in-the-mean to bound it (doing the y integral first)
Z
Z
kf − f ∗ µ kL1 ≤ µ (z) |f (y) − f (y − z)|µ (z)dydz
Z
≤ sup kf (·) − f (· − z)kL1 µ (z)dz = sup kf (·) − f (· − z)kL1
|z|≤2
|z|≤2
Here I am really assuming f ∈ Cc0 and then passing to the
limit of an approximating sequence. This shows that f ∗µ → f
in L1 .
(9) So, now we have the injectivity L1 (Rn ) ,→ S 0 (Rn ). Indeed, if
Uf = 0 then f ∗ µ = 0 for all > 0 so f = 0 in L1 .
(10) These injections allow us to regard Lp (Rn ) ⊂ S 0 (Rn ) as subspaces. In particular it makes sense to say that a tempered
LECTURE 6, 18.155, 27 SEPTEMBER 2011
5
distibution is equal to an Lp function – since this just means it
is in the range of this embedding.
(11) We use the embedding of S(Rn ) to extend the definition of differentiation to S 0 (Rn ). Indeed, if φ, ψ ∈ S(Rn ) then integration
by parts is certainly possible so
Z
Z
(∂j ψ(x))φ(x)dx = ψ(x)(−∂j φ(x))dx.
Using the notation for the identification with distributions this
reads
U∂ψ (φ) = Uψ (−∂j φ) ∀ φ ∈ S(Rn ).
But we would want the ‘distribution’ ∂j ψ, i.e. U∂ψ to be equal
to the ‘distribution’ ∂j Uψ so we just define it to be so.
Definition 1. If u ∈ S 0 (Rn ) then its partial derivatives are defined by
∂j u(φ) = u(−∂j φ) ∀ φ ∈ S(Rn ).
The continuity of ∂j : S(Rn ) −→ S(Rn ) means that the composite map on the right, −u ◦ ∂j : S(Rn ) −→ C is continuous
so this does indeed define a distribution and we have a linear
map for each j :
∂j : S 0 (Rn ) −→ S 0 (Rn ).
This is, by definition, consistent with differentiation on S(Rn )
and the injection
S(Rn )
U·
S 0 (Rn )
∂j
∂j
/
/
S(Rn )
U·
S 0 (Rn )
adding more support to our intention to regard this map as an
identification.
(12) Iterated derivatives of course behave sensibly
∂ α u(φ) = u((−1)|α| ∂ α φ).
(13) The fact that Lp functions are ‘well-defined’ distributions allows us to define the Sobolev spaces. For the moment I will
concentrate on the L2 -based spaces defined for the moment for
non-negative integral k
H k (Rn ) = {f ∈ L2 (Rn ); ∂ α Uf = fα ∈ L2 (Rn ) ∀ |α| ≤ k}.
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LECTURE 6, 18.155, 27 SEPTEMBER 2011
Notice that this ‘weak notion of derivative’ makes sense precisely because all derivatives of a tempered distribution are tempered distributions and then we ‘know’ when such a distribution
is equal to an L2 function. Observe that then by definition
(2)
∂ α Uf (φ) = fα ((−1)|α| ∂ α φ).
(14) For 1 ≤ p < ∞ we can do precisely the same thing, only changing the notation:
Lp,k (Rn ) = W p,k (Rn ) = {f ∈ Lp (Rn ); ∂ α Uf = fα ∈ Lp (Rn ) ∀ |α| ≤ k}.
The conventions for naming such spaces vary rather a lot in the
literature.