L^p spaces essentials

In one sentence

Which $L^p$ contains which depends on the measure space and on nothing else — nested one way when the measure is finite, not nested at all on $\mathbb{R}$, and nested the other way for sequences.

Why the exam asks it

Because a candidate remembers "$L^2 \subseteq L^1$" as a fact, when it is a fact about finite measure spaces only, and because $\ell^p$ inverts it. Three settings, three different answers, and the option will not tell you which setting you are in — the measure space in the stem does.

The idea

For $1 \le p < \infty$,

$$\|f\|_p = \left(\int |f|^p\right)^{1/p},$$

and $\|f\|_\infty$ is the essential supremum — the smallest $M$ with $|f| \le M$ almost everywhere. Elements are equivalence classes modulo equality a.e., which is what makes $\|f\|_p = 0 \Rightarrow f = 0$ true as a statement about the space.

Hölder: $\int |fg| \le \|f\|_p\|g\|_q$ with $\tfrac1p + \tfrac1q = 1$. Cauchy–Schwarz is $p = q = 2$.

Minkowski: $\|f + g\|_p \le \|f\|_p + \|g\|_p$ — the triangle inequality, and the reason $\|\cdot\|_p$ is a norm at all.

Riesz–Fischer: $L^p$ is complete for every $p \in [1,\infty]$. $L^2$ is the only one that is a Hilbert space, because it is the only $p$ whose norm satisfies the parallelogram law.

The inclusions, all three cases

Finite measure — say $(0,1)$. If $p < q$ then $L^q \subseteq L^p$. Hölder does it: a function that is large on a small set is controlled, and there is no room at infinity. Bigger exponent, smaller space. The inclusion is strict: $x^{-1/2}$ is in $L^1(0,1)$, since $\int_0^1 x^{-1/2} = 2$, and not in $L^2(0,1)$, since $\int_0^1 x^{-1} = \infty$.

Infinite measure — $\mathbb{R}$. Neither inclusion holds. The two witnesses:

• $x^{-1/2}\mathbf{1}_{(0,1)}$ is in $L^1$ and not $L^2$: bad near zero, where a large exponent hurts.
• $\tfrac1x \mathbf{1}_{(1,\infty)}$ is in $L^2$ and not $L^1$: $\int_1^\infty x^{-2} = 1$ while $\int_1^\infty x^{-1} = \infty$. Bad at infinity, where a large exponent helps.

The two failure modes are at opposite ends, which is exactly why finite measure rescues one of them and not the other.

Sequences — $\ell^p$, counting measure. Here $p < q$ gives $\ell^p \subseteq \ell^q$, the reverse of the finite-measure case. The reason is that there is no "near zero" problem: a sequence in $\ell^p$ has terms tending to $0$, so eventually $|a_n| < 1$ and then $|a_n|^q \le |a_n|^p$. Strictness: $(1/n) \in \ell^2 \setminus \ell^1$.

Remembering the direction: in $\ell^p$ the danger is only at infinity, and raising the exponent helps there. On a finite measure the danger is only local blow-up, and raising the exponent hurts there.

Convergence

• $L^p$ convergence does not imply pointwise convergence anywhere. The typewriter sequence — indicators of $[k/2^n, (k+1)/2^n]$ marching across $[0,1]$ — has $\|f_n\|_p \to 0$ while $f_n(x)$ fails to converge for every single $x$.
• It does imply that some subsequence converges a.e. That is the correct repair, and it is the form the true option takes.
• Pointwise convergence does not imply $L^p$ convergence either — the spike $n\mathbf{1}_{(0,1/n]}$ again.

Duality and density

For $1 \le p < \infty$ (and $\sigma$-finite measure) the dual of $L^p$ is $L^q$. The case $p = \infty$ is the exception: $L^1$ is not the dual of $L^\infty$.

Continuous functions with compact support are dense in $L^p$ for $p < \infty$, and not in $L^\infty$ — a uniform limit of continuous functions is continuous, so the closure cannot be everything. That $p < \infty$ qualifier is examined.

Where intuition breaks

"$L^2 \subseteq L^1$." True on a finite measure space, false on $\mathbb{R}$. $1/x$ on $(1,\infty)$ is the witness. The single most common error in the topic, because the finite-measure version is what gets taught first.

"$\ell^2 \subseteq \ell^1$." Backwards: $\ell^1 \subseteq \ell^2$, and $(1/n)$ shows it is strict. The $L^p$ intuition inverted, which is why the paper likes asking both in one question.

"$L^p$ convergence implies a.e. convergence." The typewriter sequence. The true statement is about a subsequence, and options trade on dropping that word.

"$L^p$ is a Hilbert space." Only $p = 2$. All of them are Banach; only one has an inner product.

"$f \in L^p$ for all $p$, so $f \in L^\infty$." False. On $(0,1)$, $f(x) = \log(1/x)$ lies in every $L^p$ with $p < \infty$ and is unbounded, so it is not in $L^\infty$.

"$\|f\|_p = 0$, so $f$ is identically zero." Only almost everywhere — and in $L^p$ that is zero, because the elements are equivalence classes. Worth being precise about, since options exploit the difference between the function and its class.

"Continuous functions are dense in $L^\infty$." No; only for $p < \infty$.

"Hölder needs $p = q = 2$." Only Cauchy–Schwarz does. Hölder holds for any conjugate pair, including the $(1,\infty)$ pair.

The exam's angle

1. Read the measure space before anything else. Finite, infinite, or counting — that decides every inclusion question in the topic, and nothing else does.
2. Keep the two witnesses ready: $x^{-1/2}$ on $(0,1)$ for $L^1 \setminus L^2$, and $1/x$ on $(1,\infty)$ for $L^2 \setminus L^1$. Between them they settle any inclusion claim on $\mathbb{R}$.
3. For $\ell^p$, reverse the direction and check with $(1/n)$.
4. On convergence options, look for the word "subsequence". Its presence usually makes a false statement true.
5. For norm inequalities, name the tool: Hölder for products, Minkowski for sums. Most computational questions here are one application of one of the two.

The night before

• Finite measure, $p < q$: $L^q \subseteq L^p$. Bigger exponent, smaller space.
• On $\mathbb{R}$: no inclusion either way. $x^{-1/2}$ on $(0,1)$ is $L^1\setminus L^2$; $1/x$ on $(1,\infty)$ is $L^2 \setminus L^1$.
• $\ell^p$: $p < q$ gives $\ell^p \subseteq \ell^q$ — the other way. $(1/n) \in \ell^2 \setminus \ell^1$.
• Hölder with conjugate exponents; Cauchy–Schwarz is the $p=q=2$ case; Minkowski is the triangle inequality.
• All $L^p$ complete (Riesz–Fischer). Only $L^2$ is Hilbert.
• $L^p$ convergence $\Rightarrow$ a.e. convergence along a subsequence only. Typewriter sequence.
• Dual of $L^p$ is $L^q$ for $p < \infty$. $L^1$ is not the dual of $L^\infty$.
• Continuous compactly supported functions dense in $L^p$ for $p < \infty$, not in $L^\infty$.
• $\log(1/x)$ on $(0,1)$: in every $L^p$, $p<\infty$, not in $L^\infty$.