Statistics Graduate Level Quiz: Convergence In Lp

15 questions · exam conditions

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Convergence In LpQuestion 1 of 15

If $X_n \to X$ in $L^p$ on any measure space, which is forced?

Convergence in measure

Almost sure convergence

Convergence in

L^q

q>p

Convergence in

L^r

r<p

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Convergence In Lp

Practice Convergence In Lp in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Convergence In Lp, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If $X_n \to X$ in $L^p$ on any measure space, which is forced?

Convergence in measure (correct answer)
Almost sure convergence
Convergence in $L^q$ , $q>p$
Convergence in $L^r$ , $r<p$

Explanation: By Markov's inequality, the measure of the set where |X_n - X| > epsilon is at most ||X_n - X||_p^p / epsilon^p, which goes to 0, so convergence in measure is forced. The tempting wrong answer is almost sure convergence: L^p convergence only guarantees a subsequence converging almost surely, not the whole sequence. L^r for r<p also fails on infinite measure spaces because L^p does not imply L^r.

Question 2

If $X_n$ is Cauchy in $L^p$ and $X_n\to X$ a.e., then

Converges only in probability
X need not even be in $L^p$
Converges to X in $L^p$ (correct answer)
May fail to converge in $L^p$

Explanation: Because L^p is complete, a Cauchy sequence has an L^p limit Y. The a.e. limit X must equal Y almost everywhere, so X is in L^p and X_n converges to X in L^p. The tempting wrong view is that a.e. convergence alone doesn't force L^p convergence; but here the Cauchy condition supplies the needed convergence.

Question 3

On a probability space, $X_n\to X$ in $L^4$ . Which statement is false?

It also converges in $L^2$
It also converges in $L^6$ (correct answer)
It converges in probability
A subsequence converges a.s.

Explanation: On a probability space, L4 convergence implies Lq convergence for every q < 4 by Lyapunov's inequality, so L2 follows and L6 is not guaranteed. It also implies convergence in probability, and convergence in probability always gives a subsequence that converges almost surely. So the false statement is the L6 claim; the tempting mistake is thinking the almost-sure subsequence result is false, but it follows from convergence in probability.

Question 4

On [0,1], let $X_n=\sqrt{n}$ on $[0,1/n]$ , else $0$ . Then

Converges in $L^1$ , not $L^2$ (correct answer)
Converges in $L^2$ , not $L^1$
Converges in $L^1$ and $L^2$
Converges a.e., not in $L^1$

Explanation: For each x>0, once n>1/x you get X_n(x)=0, so X_n converges to 0 a.e. The L1 norm is sqrt(n)(1/n)=1/sqrt(n), which tends to 0, so it converges in L1. The L2 norm is n(1/n)=1, so it does not converge in L2. The tempting wrong answer is 'converges a.e. but not in L1'; the spike at 0 grows, but that single point has measure zero and the area under the spike still vanishes.

Question 5

On a finite measure space, convergence in measure plus which condition is equivalent to $L^1$ convergence?

Finite second moments
Almost sure convergence
$L^p$ boundedness for $p>1$
Uniform integrability (correct answer)

Explanation: Convergence in measure ensures the sequence is close to the limit except on small sets; uniform integrability ensures no mass escapes to infinity on those small sets. Together they imply L1 convergence, and if L1 convergence holds, both conditions follow. L^p boundedness for p>1 is tempting because on a finite measure space it implies uniform integrability, but it is only sufficient: L1 convergence does not require any uniform L^p bound.

Question 6

Suppose $X_n\to X$ almost surely. A researcher wants to conclude that $X_n\to X$ in $L^2$ . Which additional condition is sufficient as stated?

$E[|X_n-X|]\to 0$ , since $L^1$ convergence of the differences implies $L^2$ convergence by Jensen's inequality applied to the squared error.
There exists an integrable random variable $Y$ such that $|X_n-X|^2\le Y$ almost surely for all $n$ , permitting an application of the dominated convergence theorem. (correct answer)
$\sup_n E[X_n^2]<\infty$ , since a finite uniform bound on second moments implies that the squared differences are uniformly integrable.
$P(|X_n-X|>1)\to 0$ , since controlling the probability of deviations beyond a fixed threshold is sufficient to establish mean-square convergence.

Explanation: When bridging almost sure convergence to

L^2

convergence, your go-to tool is the Dominated Convergence Theorem (DCT). Almost sure convergence tells you pointwise behavior; to upgrade to

L^2

, you need to control the integrals of

|X_n - X|^2

, which requires an integrable dominating function. Choice B is correct because if

|X_n - X|^2 \le Y

a.s. for all

n

with

E[Y] < \infty

, then since

X_n \to X

a.s. implies

|X_n - X|^2 \to 0

a.s., the DCT directly gives

E[|X_n - X|^2] \to 0

, which is exactly

L^2

convergence. Choice A is backwards and false. Jensen's inequality says

E[|X_n - X|]^2 \le E[|X_n - X|^2]

, so

L^1

convergence of differences is weaker than

L^2

, not stronger. You cannot conclude

L^2

from

L^1

. Choice C is a common trap. A uniform bound

\sup_n E[X_n^2] < \infty

does not imply uniform integrability of

|X_n - X|^2

, nor does it alone yield

L^2

convergence — you need tightness conditions beyond just bounded second moments. Choice D fails because controlling tail probabilities at a single threshold gives you convergence in probability, not

L^2

convergence. Moment convergence requires controlling the size of deviations, not merely their probability. Study tip: Whenever almost sure convergence appears and you need

L^p

convergence, immediately ask: "Can I find an integrable dominating function?" That's your DCT checklist.

Question 7

Suppose $X_n\to X$ in $L^1$ and $\sup_n\|X_n\|_3<\infty$ . Assume also that $X\in L^3$ . Which conclusion is justified?

No stronger conclusion than $L^1$ convergence is possible, because convergence in a lower norm never affects a higher norm.
$X_n\to X$ in $L^3$ , because boundedness in $L^3$ combined with $L^1$ convergence forces norm convergence.
$X_n\to X$ in essential supremum norm, because simultaneous first- and third-moment control bounds all observations.
$X_n\to X$ in $L^2$ , by interpolating between the vanishing $L^1$ distance and the bounded $L^3$ distance. (correct answer)

Explanation: When you see a problem combining

L^p

convergence with uniform boundedness in a higher norm, your instinct should be interpolation inequalities. The key tool here is the

L^p

interpolation inequality: for

1 \leq p \leq q \leq r

, we have

\|f\|_q \leq \|f\|_p^{\lambda} \|f\|_r^{1-\lambda}

where

\lambda \in [0,1]

satisfies

\frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r}

. Applying this to

X_n - X

with

p=1

q=2

r=3

, you get

\lambda = 1/2

and therefore

\|X_n - X\|_2 \leq \|X_n - X\|_1^{1/2} \cdot \|X_n - X\|_3^{1/2}.

Since

X_n \to X

L^1

, the first factor vanishes. Since

\sup_n \|X_n\|_3 < \infty

and

X \in L^3

, the quantity

\|X_n - X\|_3

is uniformly bounded. A product of something going to zero and something staying bounded goes to zero — confirming D is correct. A is wrong because it inverts the actual relationship: bounded higher norms absolutely do constrain intermediate-norm convergence through interpolation. B overclaims;

L^1

convergence plus bounded

L^3

norm is not enough to force

L^3

convergence — you'd need tightness or uniform integrability of

|X_n|^3

. C is wrong because

L^\infty

control requires far more than moment bounds; no finite moment condition, however large, implies essential supremum control. A good study habit: whenever you see "

L^p

convergence + bounded

L^r

norm," immediately write down the interpolation inequality. It almost always bridges to an intermediate

L^q

conclusion.

Question 8

Let $F_1\subseteq F_2\subseteq\cdots$ be increasing sigma-fields, let $F_\infty$ be the sigma-field generated by their union, and let $X\in L^2$ . Define $M_n=E[X\mid F_n]$ . Which statement is correct?

$M_n\to E[X]$ in $L^2$ whenever the sigma-fields are increasing, because conditioning averages out information.
$M_n\to X$ in $L^2$ regardless of whether $X$ is measurable with respect to $F_\infty$ .
$M_n$ converges only in probability unless the sigma-fields become constant after finitely many terms.
$M_n\to E[X\mid F_\infty]$ in $L^2$ , because these conditional expectations are nested orthogonal projections. (correct answer)

Explanation: When you see a question involving conditional expectations built from a filtration, think immediately about Hilbert space geometry. Conditional expectation

E[X \mid \mathcal{F}_n]

is the orthogonal projection of

X

onto the closed subspace

L^2(\mathcal{F}_n)

. As

\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \cdots

, these subspaces are nested and increasing, so the projections converge to the projection onto their limiting subspace — which is precisely

L^2(\mathcal{F}_\infty)

, where

\mathcal{F}_\infty = \sigma\!\left(\bigcup_n \mathcal{F}_n\right)

. This is the Lévy Upward Theorem:

M_n \to E[X \mid \mathcal{F}_\infty]

both almost surely and in

L^2

for any

X \in L^2

. The nested projection structure guarantees

\|M_n - E[X \mid \mathcal{F}_\infty]\|_2 \to 0

, confirming D is correct. Choice A is wrong because conditioning on more information doesn't average it away — that logic applies to decreasing sigma-fields (the Lévy Downward Theorem), where you'd converge toward

E[X \mid \mathcal{F}_\infty]

for the intersection. Increasing filtrations move toward

X

, not toward

E[X]

. Choice B nearly has the right spirit but overclaims:

M_n \to X

L^2

only if

X \in L^2(\mathcal{F}_\infty)

, i.e.,

X

is already measurable with respect to the limiting sigma-field. If

X

has components outside

\mathcal{F}_\infty

, those are irretrievably lost. Choice C invents a false condition.

L^2

convergence holds generally for any increasing filtration — no "eventually constant" requirement exists. Study tip: Always pair the direction (increasing vs. decreasing filtration) with the correct limit (

E[X \mid \mathcal{F}_\infty]

upward,

E[X \mid \mathcal{F}_\infty]

for the intersection downward). The orthogonal projection interpretation makes both theorems intuitive and memorable.

Question 9

Suppose $X_n\to X$ in $L^p$ for some finite $p\ge 1$ . Which statement about other modes of convergence is guaranteed?

The entire sequence converges almost surely, and every subsequence has the same deterministic convergence rate.
The sequence converges in probability, and at least one subsequence converges almost surely to $X$ . (correct answer)
The sequence converges in essential supremum norm after deletion of finitely many terms.
The sequence need not converge in probability, although its expectations converge whenever they are finite.

Explanation: When working with modes of convergence in probability theory, you need to know the precise relationships between

L^p

convergence, convergence in probability, and almost sure convergence — a classic hierarchy tested at the graduate level.

L^p

convergence (meaning

E[|X_n - X|^p] \to 0

) implies convergence in probability by Markov's inequality: for any

\varepsilon > 0

P(|X_n - X| > \varepsilon) \leq \varepsilon^{-p} E[|X_n - X|^p] \to 0

. However,

L^p

convergence does not generally imply almost sure convergence for the full sequence. What it does guarantee is that some subsequence converges almost surely — this follows from the fact that convergence in probability implies existence of an a.s.-convergent subsequence (a standard theorem). This makes B correct: the sequence converges in probability, and at least one subsequence converges almost surely to

X

. A is wrong on two counts:

L^p

convergence does not guarantee almost sure convergence of the entire sequence, and there is certainly no universal deterministic convergence rate across all subsequences. C is false because

L^p

says nothing about the essential supremum (

L^\infty

) norm, and convergence in

L^p

for finite

p

can fail in

L^\infty

even after removing finitely many terms. D is directly contradicted by the Markov's inequality argument above —

L^p

convergence always implies convergence in probability. As a study tip, memorize the strict hierarchy:

L^p \Rightarrow

in probability

\Rightarrow

subsequence a.s., with none of these implications reversing in general. Questions will often test whether you conflate the full-sequence and subsequence results.

Question 10

Let $U$ be uniformly distributed on $[0,1]$ , and define $X_n=n^{1/2}1_{\{U\le 1/n\}}$ . Which statement correctly describes the convergence of $X_n$ ?

$X_n \to 0$ in every finite $L^p$ space because the event on which $X_n$ is nonzero has probability tending to zero.
$X_n \to 0$ in $L^p$ exactly when $1\le p<2$ ; it also converges almost surely to zero. (correct answer)
$X_n \to 0$ in $L^2$ but not in $L^1$ , because the height of the spike grows with $n$ .
$X_n \to 0$ almost surely but fails to converge in probability because the sequence is not uniformly bounded.

Explanation: When you encounter a "spike sequence" like this one, your instinct should be to compute

\|X_n\|_p

directly rather than rely on intuition about small probabilities alone — those two factors (height and probability) interact in a way that depends critically on

p

. Here

X_n = n^{1/2} \cdot \mathbf{1}_{\{U \le 1/n\}}

, so:

E[|X_n|^p] = (n^{1/2})^p \cdot \frac{1}{n} = n^{p/2 - 1}

This expression tends to zero if and only if

p/2 - 1 < 0

, i.e.,

p < 2

. So

X_n \to 0

L^p

for all

1 \le p < 2

, and it fails to converge in

L^2

(where the norm stays at 1 for all

n

) and in any higher

L^p

. For almost sure convergence: for any fixed

\omega

X_n(\omega) \ne 0

only when

U(\omega) \le 1/n

. Since

\sum_n P(U \le 1/n) = \sum_n 1/n = \infty

, Borel–Cantelli doesn't give a.s. convergence directly — but these events are nested (decreasing), so

P(X_n \ne 0 \text{ i.o.}) = \lim P(U \le 1/n) = 0

, confirming

X_n \to 0

a.s. This confirms B is correct. A is wrong because it assumes small probability alone guarantees

L^p

convergence for all

p

— it ignores how the spike height grows. C reverses the truth:

X_n

converges in

L^1

but not in

L^2

. D is wrong on two counts: a.s. convergence holds (correct), but

L^p

convergence for

p < 2

implies convergence in probability, so the probability convergence claim is false. Your strategy: always compute

n^{p/2-1}

for spike sequences — the threshold

p = 2

is a classic exam trap.

Question 11

Fix $p\ge 1$ . Suppose $X_n \to X$ in probability and the family $\{|X_n|^p:n\ge 1\}$ is uniformly integrable. Which conclusion follows?

$X_n \to X$ in $L^p$ , because uniform integrability upgrades the probabilistic convergence of the powered errors. (correct answer)
$X_n \to X$ almost surely along the entire sequence, because uniform integrability rules out exceptional sample paths.
$X_n \to X$ in $L^r$ for every $r>p$ , because uniformly integrable variables have uniformly bounded higher moments.
$X_n \to X$ in essential supremum norm, because uniform integrability excludes arbitrarily large observations.

Explanation: When you see uniform integrability (UI) paired with convergence in probability, you should immediately think about the Vitali Convergence Theorem: convergence in probability plus uniform integrability of

\{|X_n|^p\}

is precisely the condition that upgrades to

L^p

convergence. The key mechanism is that UI controls the "tail mass" of

|X_n|^p

, preventing probability mass from escaping to infinity and spoiling the integral. More precisely, since

X_n \to X

in probability implies

|X_n - X|^p \to 0

in probability, and UI ensures the tails of

|X_n|^p

are uniformly small, we can exchange the limit and expectation to get

\mathbb{E}[|X_n - X|^p] \to 0

. This makes A correct. B is wrong because convergence in probability does not imply almost sure convergence along the entire sequence — you can only extract an a.s.-convergent subsequence. UI does nothing to fix this gap; it controls integrals, not sample-path behavior. C is wrong because UI of

\{|X_n|^p\}

does not grant uniform boundedness of higher moments. In fact, the

r

-th moments (

r > p

) may be infinite. UI only controls the

p

-th moment integrals, so

L^r

convergence for

r > p

is not guaranteed. D is wrong because UI is an

L^1

-type integrability condition on tails — it says nothing about essential supremum (

L^\infty

) control. The

X_n

can still be unbounded a.s. Study tip: Memorize the Vitali theorem as a clean package: in probability + UI of

|X_n|^p

\Rightarrow

L^p

convergence. This trio appears repeatedly on graduate probability exams.

Question 12

On the measure space consisting of the real line with Lebesgue measure, define $f_n(x)=n^{-1}1_{[0,n]}(x)$ . Which statement is correct?

$f_n\to 0$ in both $L^1$ and $L^2$ , because the pointwise height tends to zero.
$f_n\to 0$ in $L^1$ but not in $L^2$ , because the supports have unbounded length.
$f_n\to 0$ in $L^2$ but not in $L^1$ , illustrating that finite-measure norm inclusions can fail here. (correct answer)
$f_n$ fails to converge to zero in either norm, because each function has support of increasing measure.

Explanation: When working with

L^p

convergence on infinite measure spaces like

(\mathbb{R}, \mathcal{B}, \lambda)

, you cannot rely on intuition built from finite-measure settings. The key is to compute the norms directly rather than reason qualitatively about "height" or "support size" alone. For

f_n(x) = n^{-1}\mathbf{1}_{[0,n]}(x)

, compute each norm explicitly. The

L^1

norm is

\|f_n\|_1 = \int_0^n n^{-1}\,dx = n^{-1} \cdot n = 1,

which stays constant at 1 for all

n

, so

f_n \not\to 0

L^1

. The

L^2

norm is

\|f_n\|_2 = \left(\int_0^n n^{-2}\,dx\right)^{1/2} = \left(n^{-2} \cdot n\right)^{1/2} = n^{-1/2} \to 0,

confirming

f_n \to 0

L^2

. This makes C correct. The intuition here is that on infinite measure spaces,

L^1 \not\subset L^2

and

L^2 \not\subset L^1

. This sequence exploits exactly that gap — the shrinking height decays fast enough for

L^2

(which penalizes large values more heavily) but not for

L^1

, because the expanding support perfectly compensates the shrinking height. A is wrong because it incorrectly concludes

L^1

convergence from pointwise decay — pointwise convergence to zero never guarantees

L^1

convergence without uniform integrability. B gets the

L^1

conclusion backward — the constant

L^1

norm means it fails in

L^1

, not converges. D is wrong because the

L^2

norm calculation clearly shows convergence to zero. Your strategy: always compute norms explicitly before concluding anything about

L^p

convergence, especially on infinite measure spaces where norm inclusions from the finite setting no longer hold.

Question 13

Let $1\le p<\infty$ . Assume that $\{X_n\}$ is Cauchy in $L^p$ and that $X_n\to X$ in probability for some random variable $X$ . What must be true?

The sequence converges to $X$ in $L^p$ , by completeness of $L^p$ and uniqueness of limits in probability. (correct answer)
The sequence converges to $X$ almost surely, because every $L^p$ -Cauchy sequence converges pointwise.
The sequence converges in $L^p$ to some variable, but that variable may differ from $X$ on an event of positive probability.
No $L^p$ conclusion follows unless $\sup_n|X_n|$ is dominated by an integrable random variable.

Explanation: When you see a question combining an

L^p

-Cauchy sequence with convergence in probability, you should immediately think about two foundational theorems: the completeness of $L^p$ and the uniqueness of limits in probability. Here's the chain of reasoning. Since

\{X_n\}

is Cauchy in

L^p

and

L^p

is a complete metric space (for

1 \le p < \infty

), the sequence must converge in

L^p

to some limit, call it

Y

. Now, convergence in

L^p

implies convergence in probability (by Markov's inequality). So

X_n \to Y

in probability. But we're also told

X_n \to X

in probability. Since limits in probability are unique up to almost-sure equality, we conclude

X = Y

a.s. Therefore

X_n \to X

L^p

, confirming A is correct. B is wrong because

L^p

-Cauchy sequences do not generally converge pointwise or almost surely — convergence in

L^p

is a statement about norms, not trajectories. Classic counterexamples (the "typewriter sequence") show

L^p

convergence without almost-sure convergence. C is wrong because it contradicts the uniqueness of limits in probability. Two

L^p

limits that differ on a positive-probability event would define different probability limits, which is impossible. D is wrong because domination by an integrable function is sufficient for applying the Dominated Convergence Theorem, but it is not necessary here. Completeness of

L^p

alone guarantees the conclusion without any domination condition. Study tip: Memorize this chain — Cauchy in

L^p

\Rightarrow

L^p

limit exists (completeness)

\Rightarrow

probability limit exists

\Rightarrow

both probability limits must agree a.s. This three-step argument appears frequently on graduate probability exams.

Question 14

Let $1 \le p < q < \infty$ , and suppose the underlying probability space has total mass one. If $X_n \to X$ in $L^q$ , which conclusion is necessarily valid without any additional assumptions?

$X_n \to X$ in $L^p$ , because the $L^p$ distance is bounded by the $L^q$ distance on a probability space. (correct answer)
$X_n \to X$ almost surely, because convergence of a higher-order norm controls every sample path.
$X_n \to X$ in $L^r$ for every finite $r>q$ , because the sequence is bounded in $L^q$ .
$X_n \to X$ in essential supremum norm, because finite probability prevents large tail deviations.

Explanation: Whenever you see a question mixing

L^p

spaces on a probability space (total mass one), your first instinct should be the norm comparison inequality: for

1 \le p < q < \infty

and a probability measure, Jensen's inequality gives

\|Y\|_p \le \|Y\|_q

. This is the heart of this problem. Applying this to

Y = X_n - X

, you get

\|X_n - X\|_p \le \|X_n - X\|_q

. So if

\|X_n - X\|_q \to 0

, it immediately follows that

\|X_n - X\|_p \to 0

. The

L^q

norm controls the

L^p

norm from above when

p < q

and the measure is a probability measure — this is why A is correct. B is wrong because

L^q

convergence implies only convergence in probability (and hence a subsequence converges a.s.), but not a.s. convergence of the full sequence. A classic counterexample is the "typewriter sequence" of indicator functions, which converges in every

L^p

but nowhere almost surely. C reverses the direction of the inequality. Being bounded in

L^q

tells you nothing about finiteness of

L^r

norms for

r > q

; higher moments can be infinite even if lower ones are finite. D is false because

L^\infty

(essential supremum) is the largest

L^p

space — it is not controlled by finite

L^q

norms. The essential supremum can be infinite even when all finite-order moments exist. Study tip: On probability spaces, the norm inequality

\|Y\|_p \le \|Y\|_q

for

p < q

flows downward — stronger (higher) norms control weaker (lower) norms. Keep this direction firmly in mind; many distractors exploit reversing it.

Question 15

Suppose $X_n\to X$ in $L^2$ and $Y_n\to Y$ in $L^2$ . Which assertion about the products is necessarily correct?

$X_nY_n\to XY$ almost surely along the entire sequence, because both factors converge in mean square.
$X_nY_n\to XY$ in $L^2$ , because multiplication preserves convergence in any fixed $L^p$ space.
$X_nY_n\to XY$ in $L^1$ , by decomposing the product error and applying the Cauchy–Schwarz inequality. (correct answer)
$X_nY_n\to XY$ in essential supremum norm, because both factor sequences are bounded in $L^2$ .

Explanation: When combining two convergent sequences through multiplication, you need to think carefully about which mode of convergence is preserved and why — this question tests exactly that interplay between

L^2

convergence and product stability. The key technique is the algebraic decomposition: write

X_nY_n - XY = (X_n - X)Y_n + X(Y_n - Y)

. Now apply the Cauchy–Schwarz inequality to each term. For the first:

\|{(X_n-X)Y_n}\|_1 \leq \|X_n - X\|_2\|Y_n\|_2

. Since

X_n \to X

L^2

, the first factor vanishes; since

Y_n \to Y

L^2

, the sequence

(Y_n)

is bounded in

L^2

. The second term

\|X(Y_n - Y)\|_1 \leq \|X\|_2\|Y_n - Y\|_2 \to 0

similarly. Both terms go to zero, so

X_nY_n \to XY

L^1

— confirming C is correct. A is wrong because

L^2

convergence implies only convergence in probability (and thus a.s. along a subsequence), not almost sure convergence along the entire sequence. Claiming a.s. convergence for the full sequence is an unjustified leap. B fails because multiplication does not generally preserve

L^2

convergence. The product

X_nY_n

need not even be in

L^2

without additional assumptions like uniform boundedness, and the bound you'd need requires

L^4

control, not

L^2

. D is false because

L^2

boundedness says nothing about essential supremum (

L^\infty

) control; the sequences could be unbounded pointwise. Your takeaway: whenever you see a product of two

L^2

-convergent sequences, reach immediately for the decomposition

X_nY_n - XY = (X_n-X)Y_n + X(Y_n-Y)

paired with Cauchy–Schwarz — it's the canonical tool for dropping to

L^1