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$L^p$ Spaces

02 Sep 2021 - Tags: analysis-qual-prep

On to day $2$ of qual prep boogaloo. Yesterday Tuesday1 was all about measures and integrable functions. Today, then, let’s talk about spaces of integrable functions. It’s time for $L^p$ spaces!

Thankfully, this should be a comparatively short post, since $L^p$ spaces are already pretty concrete, and (at least in my mind) don’t need a ton of motivation. We want to integrate things, and $L^p$ spaces (for $p \lt \infty$) collect the functions which we can integrate.

The lower values of $p$, such as $p=1$, put emphasis on the decay of a function near $\infty$. The benefit we gain is that they place relatively little emphasis on singularities.

Conversely, large values of $p$, such as $p=\infty$, put little emphasis on the decay of a function, but are more sensitive to (potentially mild) singularities.

As an example, think of $\frac{e^{-x}}{\sqrt{x}}$:

This has a singularity at $0$, but is still in $L^1(\mathbb{R})$ because it decays rapidly. This function is not in $L^2(\mathbb{R})$ (and certainly not in $L^\infty$), because its singularity is too sharp.

Conversely, consider the constant $1$ function, shown here for completeness more than anything else:

This function doesn’t decay at all, so is certainly not $L^1$. It turns out this function is only $L^\infty$, which doesn’t care at all about decay. It only cares about boundedness.

A less silly example might be $\chi_{[1,\infty)} \frac{1}{x}$:

This function decays too slowly to be $L^1$, but it does decay quickly enough to be $L^2$.

The most important cases are $L^1$, $L^2$, and $L^\infty$, but there are reasons to care about other choices of $p$ as well. Psychologically, they let us interpolate between how much singularity we want to allow, versus how slowly we want to allow our functions to decay.

To start, let’s remind ourselves what $L^p$ spaces even are. We fix a set $X$ with a measure $\mu$. Then (for $1 \leq p \lt \infty$), we define the $p$-norm to be

\[\lVert f \rVert_p \triangleq \left ( \int |f|^p \ d \mu \right )^{1/p}\]

As is often the case, we need to treat $L^\infty$ specially. We write2

\[\lVert f \rVert_\infty \triangleq \inf \Big \{ M \ \Big | \ |f| \leq M \text{ a.e.} \Big \}\]

Now we can define (for $1 \leq p \leq \infty$)

\[L^p(X, \mu) \triangleq \bigg \{ f : X \to \mathbb{C} \ \bigg | \lVert f \rVert_p \lt \infty \bigg \}\]

As usual, we will consider two functions the same if they agree almost everywhere. We will also often write $L^p(X)$, $L^p(\mu)$, or even $L^p$ if the measure or space is clear from context.

It turns out $L^p(\mu)$ comes equipped with a ton of structure! The norm $\lVert f \rVert_p$ really is a norm3, and this norm makes $L^p(\mu)$ into a banach space (so we can talk about convergence without worry4).

Let’s see some important examples of $L^p$ spaces before we move on:

  1. If $m$ is the lebesgue measure on $\mathbb{R}^n$, then $L^p(\mathbb{R}^n, m)$ is one of the most common examples used in practice. This forms the basis for a lot of intuition about $L^p$ spaces.

  2. Sometimes $\mathbb{R}^n$ is too big. We might think of working with $L^p([a,b])$. Common choices are $[0,1]$ and $[0,2\pi]$.

  3. Another important example is the circle $S^1$. The space $L^2(S^1)$ is the archetypal first example for fourier theory5.

  4. An example in a different vein is $L^p(\mathbb{N})$ with the counting measure. This is often called $\ell^p$. Here our functions can be thought of as sequences of complex numbers, which makes this a particularly appealing setting for small examples. We’ll find $\ell^2$ extremely useful when we talk about hilbert spaces.

  5. The counterpoint to example (3) is $L^p(\mathbb{Z})$, again with the counting measure. These are bi-infinite sequences of complex numbers, often written $\ell^p(\mathbb{Z})$, and these will be related to functions on the circle by means of the fourier transform.

One of the most important tools we have for proving properties about $L^p$ functions is the wide variety of dense subsets. If we want to prove something for all $L^p$ functions, we can start out by proving that thing for some subclass where we have more available tools, then argue by continuity that the result is actually true for all $L^p$ functions.

These sets of functions are dense in $L^p(\mathbb{R}^n)$ and $L^p([0,1])$, where $p \neq \infty$. More generally, they are true in $L^p(X,\mu)$ for radon measures on locally compact hausdorff spaces. See Rudin’s Real and Complex Analysis for more details.

These families of dense functions require a notion of differentiation. They’re definitely true in $\mathbb{R}^n$ and $[0,1]$, and I suspect they’re true over any manifold. I don’t have a reference on hand, though.

These families of functions also require some compactness condition (because we’re using the stone-weierstrass theorem).

There’s actually a kind of master theorem for density results in $L^p$ spaces ($p \neq \infty$), which takes some basic properties of a class of functions and tells you it’s dense. There are also generalizations of the theorems above to much more broad classes of measure spaces. For each of these results, see this pdf from Bruce Driver’s UCSD lecture notes.

Before we move on, these density results are one potential source of motivation for $L^1$. We’re obviously interested in continuous functions, and if we want to integrate things, the $L^1$ norm seems like a fairly reasonable metric to look at. The issue is that $L^1$ limits of continuous functions don’t need to be continuous! That is, $CX$ is not complete with respect to the $L^1$ norm. We like to know we can take limits with reckless abandon7, so we should take its completion. When we do this, we get exactly the $L^1$ functions.

This is one reason that $L^\infty$ behaves “pathologically” compared to the other $L^p$ spaces. When you take the completion of continuous functions with compact support in the $p$-norm for $p \lt \infty$, get get exactly $L^p$. However, when you take the completion in the $\infty$-norm, you get the functions vanishing at $\infty$. You might try to solve this by looking at the bounded continuous functions, rather than those of compact support8, but these already form a banach space! It seems that, unavoidably, $L^\infty$ has a bunch of extra stuff in it, and it’s this extra stuff that is responsible for many of the differences between $L^\infty$ and the other $L^p$ spaces.

Perhaps the most important inequality associated with $L^p$ spaces is Hölder’s inequality, which generalizes the Cauchy-Schwarz inequality.

If $\frac{1}{p} + \frac{1}{q} = 1$, then

\[\lVert fg \rVert_1 \leq \lVert f \rVert_p \lVert g \rVert_q.\]

This includes the formal case where $\frac{1}{1} + \frac{1}{\infty} = 1$:

\[\lVert fg \rVert_1 \leq \lVert f \rVert_1 \lVert g \rVert_\infty.\]

This condition on $p$ and $q$ shows up frequently, and we call $q$ the conjugate of $p$ for convenience.

Here we might think of $g$ as defining a functional $\phi_g$, where $\phi_g f \triangleq \int f g$. Then Hölder’s inequality says that (with the operator norm) $\lVert \phi_g \rVert \leq \lVert g \rVert_q$.

From this, one might ask if we actually have equality above. If you’re feeling particularly optimistic, you might even wonder if we can characterize which functionals in $(L^p)^*$ arise as $\phi_g$ for some $g \in L^q$.

Now for the magic part:

For $p \neq 1, \infty$ the dual space \((L^p)^* \triangleq \{ T : L^p \to \mathbb{C} \mid T \text{ is continuous and linear} \}\) has an explicit characterization:

\[(L^p)^* \cong L^q\]

where $q$ is the conjugate of $p$.

When $\mu$ is $\sigma$-finite9, we moreover have $(L^1)^* \cong L^\infty$. Unfortunately, though, $(L^\infty)^*$ is almost never $L^1$ (at least, assuming the axiom of choice10).

The proof of this fact actually goes through complex valued measures! If $\phi$ is a functional on $L^p$, then it’s not hard to see that $\nu E \triangleq \phi \chi_E$ is a (complex valued) measure! Moreover, if $E$ is $\mu$-null, then $\chi_E = 0$ a.e., and so $\nu E = \phi \chi_E = 0$. This means $\nu \ll \mu$, and so by radon-nikodym $\nu = \mu_g$ for some function $g$. Now some simple bookkeeping is all we need to do to show that $g \in L^q$ and $\phi f = \int fg$.

Notice this means that for $1 \lt p \lt \infty$ we have

\[(L^p)^{**} = (L^q)^* = L^p\]

so $L^p$ is reflexive. It’s still not entirely obvious to me why reflexive spaces are as interesting as they are. It’s obviously a natural question (and a fairly algebraic one11), and it gives you quite a bit of bonus information (see here, for instance). I suspect that I’ll come to appreciate it more as I do more functional analysis12. At the very least, it’s nice to have a full characterization of the dual space, and it’s satisfying to get back where you started when you dualize twice.

Frequently one has a function $f \in L^p$, which we would like to know is in $L^q$. More generally, we would like to relate functions in $L^p$ to functions in $L^q$ with $q$ on either side of $p$.

For instance, when doing fourier analysis, our hilbert space techniques only work for $L^2(S^1)$ functions, which get mapped by the fourier transform isometrically onto $\ell^2(\mathbb Z)$. However, we can show that $\ell^1(\mathbb Z) \subseteq \ell^2(\mathbb Z)$, and thus we see that if $f$ is $L^1(S^1)$ and $\hat{f}$ is $\ell^1(\mathbb Z)$, then the results (which we’ll talk about soon) still hold.

With that said, here are the main theorems for useful inclusions:

If $X$ has finite measure, then going down is allowed. That is, if $q \geq p$, then every $L^q$ function is automatically $L^p$.

Formally, if $q \geq p$, then $L^q \subseteq L^p$.

(As an easy exercise, you should prove this)

If $X$ is “discrete” in the sense that it does not have sets of arbitrarily small measure, then going up is allowed. That is, if $p \leq q$, then every $L^p$ function is automatically $L^q$.

Formally, if $p \leq q$, then $L^p \subseteq L^q$.

(Again, this is a fairly easy exercise)

In fact, both of the above statements are equivalences. So going down is allowed if and only if $X$ has finite measure, and going up is allowed if and only if $X$ has subsets of minimal measure13.

Notice how these line up with the singularity/rapid decay tradeoff we mentioned earlier. If $X$ has finite measure, then (intuitively) there’s no “infinity” to worry about decaying towards. So the decay penalization is irrelevant, and we get an inclusion of “lower singularity” functions into “higher singularity” functions with no further qualifications14.

Conversely, if there are sets of minimal measure, then (intuitively) any singularity on one of your atoms can’t be avoided, and we get the inclusion of “rapidly decaying” functions into “slowly decaying” functions with no further qualifications15.

Of course, once we know that, say $L^2([0,1]) \subseteq L^1([0,1])$, it’s natural to ask “how many $L^1$ functions are actually $L^2$?”. The answer is “almost none”!

It turns out that $L^2([0,1])$ is meagre in $L^1([0,1])$. This is a topological notion of “smallness” which is roughly akin to being a nullset16. This is a cute application of the baire category theorem, which is the topic of an upcoming blog post.

We also have interpolation functions which, if $p \lt q \lt r$, lets us relate $L^q$ functions to those on either side.

If $p \lt q \lt r \lt \infty$, then $L^q \subseteq L^p + L^r$.

That is, every $L^q$ function can be written as a sum of an $L^p$ function and an $L^r$ function.

If $p \lt q \lt r \lt \infty$, then $L^p \cap L^r \subseteq L^q$.

That is, the $L^p$ classes are “connected”. If you’re $L^p$ and you’re $L^r$, then you’re automatically everywhere in between as well.

In fact, we have the bound

\[\lVert f \rVert_q \leq \lVert f \rVert_p^\lambda \lVert f \rVert_r^{(1 - \lambda)}\]

for $\lambda = \frac{q^{-1} - r^{-1}}{p^{-1} - r^{-1}}$.

Alright, that was another huge information dump! This post felt less like a motivated tour through $L^p$ functions and more like a copy of a study guide, haha. I think that’s reasonable, though, since in my experience $L^p$ functions are pretty motivated by themselves. I’m mainly writing this so that when I talk about Banach spaces soon we’ll have all these examples to draw from. Plus, selfishly, it was nice to organize all this information while planning out this post.

I’ve learned from my mistakes, and I’m not going to try and give a concrete date for the next post. All I’ll say is: See you soon ^_^

  1. I’ve been busier than expected… So much for one post a day, haha. 

  2. The notation comes from a (non-obvious) theorem that

    \[\lVert f \rVert_\infty = \lim_{p \to \infty} \lVert f \rVert_p.\]

    See here for a proof. 

  3. This is one pressing reason to quotient by almost-everywhere equivalence. Do you see why $\lVert f \rVert_p$ would not be a norm if we didn’t identify equivalent functions?

    One reasonable question might be whether we can always find a canonical representative for a given equivalence class modulo nullsets. That is, given a “function” in $L^p$ (which is really an equivalence class of functions) can we select an honest-to-goodness function from each class? Obviously the axiom of choice trivializes this, but we would like to do so in a way that’s actually implementable.

    This is an extremely interesting question, and is the fundamental question in lifting theory

  4. Don’t worry – we’ll be talking a lot about banach spaces in an upcoming post. But $L^p$ spaces are very fundamental examples, so it’s worth talking about them first. 

  5. We’ll talk about Fourier theory in an upcoming post too! 

  6. Recall we define the lebesgue integral of $f$ to be the limit of the integrals of simple functions below $f$. 

  7. This is in line with an apocryphal quote of Grothendieck (which is probably not actually his) that it’s better to have good categories with bad objects, than bad categories with good objects.

    That is, we want our categories to have all limits, colimits, exponentials, etc. even if it means introducing “pathological” objects. It’s easier to know you can take limits and worry about what the result looks like later than it is to constantly worry about whether you can take the limit at all.

    Similarly, from an analytic lens, it’s better to have a complete space (which admits potentially ugly functions) than it is to have an incomplete space in which every function is nice. 

  8. You need boundedness in order to guarantee that the sup norm $\lVert \cdot \rVert_\infty$ is finitely valued. 

  9. Actually, we can get by with slightly less.

    We always have a natural map $L^\infty \to (L^1)^*$ which sends $g$ to the functional $f \mapsto \int fg$.

    It turns out this map is an isometric embedding if and only if $\mu$ is semifinite. That is, when any infinite measure set contains a set of finite measure. This is obviously a super mild assumption.

    Moreover, this map is surjective (and thus an isometry) when $\mu$ is “localizable”. See here, for instance. 

  10. $L^1$ embeds isometrically inside $(L^\infty)^*$, but in $\mathsf{ZFC}$ this is not surjective.

    To build a functional on $L^\infty$ that doesn’t come from some $L^1$ function, we look at the subspace of continuous functions, and consider evaluation at a point. We can extend this functional to the whole space via the Hahn-Banach Theorem, and it’s pretty quick to see that this can’t come from an $L^1$ function (since a dirac delta function doesn’t really exist).

    We’ll talk more about the Hahn-Banach theorem in a future post, but importantly it relies on the axiom of choice! There are actually models of set theory where Hahn-Banach fails, and which think that $L^1$ really is $(L^\infty)^*$! See Martin Väth’s The Dual Space of $L^\infty$ is $L^1$, as well as the excellent discussion here 

  11. The question of which $R$-modules are reflexive has also been asked. It’s a more subtle question, but finitely generated projective modules always have this property.

    As an aside that I can’t help but mention, apparently there is a banach space which is isomorphic to its double dual, but for which the canonical evaluation map is not the isomorphism! It’s called James’s Space, for the interested. 

  12. For instance, an integral that takes values in a banach space might not satisfy the radon-nikodym theorem. If your banach space is reflexive, though, then it does. This came from the linked mse question, so I might be misquoting the result or dropping some hypotheses, but this already makes the notion of reflexivity seem more interesting. 

  13. For proofs of these exercises, and indeed proofs of the stronger equivalences, see here 

  14. This may require some meditation.

    The idea is that we don’t care if a function is singular as long as the infinity can be contained to a set of small enough measure. But if there are no sets of “small enough” measure, then any singularity is bad. 

  15. In the motivational section before this, we mention that $\ell^1(\mathbb Z) \subseteq \ell^2(\mathbb Z)$. Now we see why: $\mathbb{Z}$ has no (nonnull) sets of measure $\lt 1$. 

  16. In the sense that meagre sets and nullsets both form $\sigma$-ideals. Note, however, that they are very different notions of “small”, and a set which is small in one sense need not be small in the other.

    For instance, fat cantor sets can have arbitrarily large measure (in particular, they are not null) yet they are always meagre.