Lower Semicontinuity of the Norm Under Weak Limits

Weak Limits and the Norm

Weak convergence tests a sequence by applying every bounded linear functional, rather than by measuring the norm of its difference from the proposed limit. Consequently, weak convergence need not mean that the sequence gets close in norm. Nevertheless, it imposes a precise restriction: the norm of a weak limit cannot exceed the lower limit of the norms along the sequence. This restriction is useful both for estimating limits and for identifying sets that are closed in the weak topology.

Throughout, let $(X,\mathcal{M},\mu)$ be any measure space, and let $1\le p\le\infty$. No finiteness, sigma-finiteness, or completeness assumption on $\mu$ is needed. The space $L^p(\mu)$ consists of equivalence classes, modulo equality $\mu$-a.e., of real- or complex-valued measurable functions with finite $L^p$ norm. For $p<\infty$, $\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}$; for $p=\infty$, $\|f\|_\infty$ is the essential supremum of $|f|$. We use the weak topology defined by all bounded linear functionals on $L^p(\mu)$.

Definition: A sequence $(f_n)$ in $L^p(\mu)$ converges weakly to $f\in L^p(\mu)$, written $f_n\rightharpoonup f$, if $\ell(f_n)\to\ell(f)$ for every bounded linear functional $\ell:L^p(\mu)\to\mathbb{R}$ (or $\mathbb{C}$ in the complex case). These are statements about equivalence classes; the convergence is not a pointwise or almost-everywhere assertion.

The proof rests on expressing the norm through bounded linear functionals. We use the Hahn–Banach theorem, a standard theorem of functional analysis, without proving it here. Its consequence needed below is established directly.

Lemma (Dual formula for the norm): For a normed space $Y$ over $\mathbb{R}$ or $\mathbb{C}$, let $Y^*$ be its bounded linear functionals, with $\|\ell\|=\sup_{\|y\|\le1}|\ell(y)|$. For every $y\in Y$, $$ \|y\|=\sup\{|\ell(y)|:\ell\in Y^*,\ \|\ell\|\le1\}. $$

Proof. For each $\ell\in Y^*$ with $\|\ell\|\le1$, the definition of the functional norm gives $|\ell(y)|\le\|y\|$. Thus the supremum on the right is at most $\|y\|$. If $y=0$, every functional vanishes at $y$, so both sides are zero. If $y\ne0$, define a functional on the one-dimensional subspace spanned by $y$ by $\ell_0(\alpha y)=\alpha\|y\|$. This is well-defined and linear, and $|\ell_0(\alpha y)|=|\alpha|\|y\|=\|\alpha y\|$, so its norm is $1$. By Hahn–Banach, $\ell_0$ extends to a functional $\ell\in Y^*$ with norm $1$. Then $|\ell(y)|=\|y\|$, proving that the supremum is at least $\|y\|$. $\square$

The Lower Semicontinuity Inequality

Theorem: Let $(X,\mathcal{M},\mu)$ be any measure space, $1\le p\le\infty$, and $(f_n)$ a sequence in $L^p(\mu)$ that converges weakly to $f\in L^p(\mu)$. Then $$ \|f\|_p\le\liminf_{n\to\infty}\|f_n\|_p. $$ The norms and the limit inferior are finite or extended nonnegative real numbers; the conclusion includes the case where the limit inferior is $+\infty$.

Proof. Write $L=\liminf_n\|f_n\|_p$. Fix any bounded linear functional $\ell$ on $L^p(\mu)$ with $\|\ell\|\le1$. Weak convergence gives $\ell(f_n)\to\ell(f)$, and continuity of absolute value gives $|\ell(f_n)|\to|\ell(f)|$. For every $n$, the definition of $\|\ell\|$ implies $|\ell(f_n)|\le\|f_n\|_p$. Taking lower limits in this inequality yields $$ |\ell(f)|=\liminf_{n\to\infty}|\ell(f_n)| \le\liminf_{n\to\infty}\|f_n\|_p=L. $$ This holds for every such $\ell$. Taking the supremum over them and applying the dual formula to $f$ gives $\|f\|_p\le L$. If $L=+\infty$, the inequality is automatic; the same argument still applies. $\square$

This proof does not identify the dual of $L^p(\mu)$ with a particular function space. That distinction matters at the endpoints: for $1<p<\infty$, the usual $L^q$ dual representation applies under the standard hypotheses of the $L^p$ duality theorem, while at $p=\infty$ the full dual need not be represented by $L^1$ functions. The argument above uses the full bounded dual in every case, so the lower semicontinuity result holds for all $1\le p\le\infty$.

Worked Example: A Weakly Null Sequence with Norm One

Take $X=\mathbb{N}=\{1,2,\ldots\}$ with counting measure, and $p=3$. Then $L^3(\mu)=\ell^3$. Let $e_n$ be the sequence whose $n$-th coordinate is $1$ and whose other coordinates are $0$. We verify that $e_n\rightharpoonup0$, even though its norms do not tend to zero.

By the $L^p$ duality theorem for $1<p<\infty$, each bounded linear functional on $\ell^3$ has the form $\ell_a(x)=\sum_{k=1}^\infty x_k a_k$ for some $a\in\ell^{3/2}$. (In the complex case, use the corresponding conjugate pairing.) For each such $a$, $\ell_a(e_n)=a_n$. Since $\sum_k|a_k|^{3/2}<\infty$, its terms satisfy $a_n\to0$: otherwise some $\varepsilon>0$ would have $|a_n|\ge\varepsilon$ for infinitely many $n$, forcing the series to diverge. Thus every bounded functional tends to zero on $e_n$, which proves $e_n\rightharpoonup0$.

There is exactly one nonzero coordinate in $e_n$, so $\|e_n\|_3=(|1|^3)^{1/3}=1$ for every $n$. Hence $\liminf_n\|e_n\|_3=1$, whereas $\|0\|_3=0$. The theorem gives $0\le1$, strictly. Check: applying the coordinate functional $x\mapsto x_j$ for a fixed $j$ gives $1$ when $n=j$ and $0$ for all $n\ne j$, consistent with weak convergence to zero.

Worked Example: Weak Convergence When the Norms Also Converge

Let $X=[0,1]$ with Lebesgue measure and $p=2$. Set $u=\chi_{[0,1/2]}$, and define $f_n=(1+1/n)u$, with $f=u$ as the proposed limit. We check weak convergence and compute the norms.

First, $\|f_n-f\|_2=(1/n)\|u\|_2$. The interval $[0,1/2]$ has Lebesgue measure $1/2$, so $\|u\|_2=(\int_0^1|u|^2\,dm)^{1/2}=(1/2)^{1/2}=1/\sqrt{2}$. Therefore $\|f_n-f\|_2=1/(n\sqrt{2})\to0$. For every bounded linear functional $\ell$ on $L^2(m)$, $$ |\ell(f_n)-\ell(f)|\le\|\ell\|\,\|f_n-f\|_2 =\frac{\|\ell\|}{n\sqrt{2}}\longrightarrow0. $$ Thus $f_n\rightharpoonup f$. Direct computation gives $$ \|f_n\|_2=(1+1/n)\|u\|_2=\frac{1+1/n}{\sqrt{2}}, \qquad \|f\|_2=\frac{1}{\sqrt{2}}. $$ The norms tend to $1/\sqrt{2}$, so the lower semicontinuity inequality is an equality. Check: for $n=1$, $\|f_1\|_2=2/\sqrt{2}=\sqrt{2}$; for $n=2$, $\|f_2\|_2=(3/2)/\sqrt{2}=3/(2\sqrt{2})$. Both agree with the formula and decrease toward $1/\sqrt{2}$.

Closed Norm Balls in the Weak Topology

Corollary: For any measure space $(X,\mathcal{M},\mu)$, $1\le p\le\infty$, and $R\ge0$, the closed ball $B_R=\{f\in L^p(\mu):\|f\|_p\le R\}$ is closed in the weak topology.

Proof. By the dual formula, $$ B_R=\bigcap_{\substack{\ell\in (L^p(\mu))^*\\\|\ell\|\le1}} \{f\in L^p(\mu):|\ell(f)|\le R\}. $$ Indeed, if $\|f\|_p\le R$, then $|\ell(f)|\le\|f\|_p\le R$ for every functional in the intersection. Conversely, if $|\ell(f)|\le R$ for every such functional, their supremum is $\|f\|_p$, so $\|f\|_p\le R$. Each set in the intersection is weakly closed: the weak topology makes $\ell$ continuous, and the closed disk $\{z:|z|\le R\}$ is closed in the scalar field. An arbitrary intersection of closed sets is closed, proving the claim. This argument also covers $R=0$ and the zero space. $\square$

Worked Example: Separating a Point Outside a Weakly Closed Ball

In $\ell^3$ with counting measure, consider the closed unit ball $B_1=\{y:\|y\|_3\le1\}$ and $x=(2,0,0,\ldots)$. We exhibit a weakly open neighborhood of $x$ that misses $B_1$.

The functional $\ell(y)=y_1$ is bounded and has norm $1$: $|y_1|\le(\sum_k|y_k|^3)^{1/3}=\|y\|_3$, and equality holds at $y=e_1$. Also $\ell(x)=2$. The set $U=\{y:|\ell(y)-2|<1/2\}$ is weakly open and contains $x$, since $|2-2|=0<1/2$. If $y\in B_1$, then $|y_1|\le1$, and the triangle inequality gives $|y_1-2|\ge 2-|y_1|\ge1$. Thus $y\notin U$. This explicitly verifies that the point outside the ball can be excluded by a weakly open set. Check: $x$ has norm $(|2|^3)^{1/3}=2>1$, as required.

A Common Misreading

A common mistake is to replace the lower semicontinuity inequality by the stronger claim that weak convergence implies convergence of norms. The first worked example disproves that claim: $e_n\rightharpoonup0$ in $\ell^3$, but $\|e_n\|_3=1$ for every $n$, while $\|0\|_3=0$. The correct conclusion is only $\|f\|_p\le\liminf_n\|f_n\|_p$; equality or convergence of norms requires additional information.

It is also important to test weak convergence against the full dual. For example, the coordinate vectors $e_n$ in $\ell^1$ do not converge weakly to zero. The bounded functional $\ell(x)=\sum_{k=1}^\infty x_k$ is well-defined on $\ell^1$, since $|\ell(x)|\le\sum_k|x_k|=\|x\|_1$. It satisfies $\ell(e_n)=1$ for every $n$, not $0$. Thus a sequence with constant norm is not, by that fact alone, a weakly null sequence. The weak-limit hypothesis must be checked before applying the theorem.

The norm formula explains the geometric content of the result: every unit-norm functional gives a scalar test of the limit, and the norm is the supremum of all such tests. A weak limit preserves each fixed test value, so none of those tests can exceed the lower limit of the sequence norms. Taking the supremum yields the inequality. The same representation makes norm balls intersections of weakly closed test sets, which is why they are weakly closed.

Check Your Understanding

For each question, distinguish the norm estimate from the definition of weak convergence.

In the dual formula for a norm, why is the supremum at most $\|y\|$, and how does Hahn–Banach provide the reverse inequality when $y\ne0$?
If $f_n\rightharpoonup f$ and $\ell$ has norm at most $1$, what inequality relates $|\ell(f)|$ to $\liminf_n\|f_n\|_p$?
Why does weak convergence of $e_n$ to zero in $\ell^3$ not imply that $\|e_n\|_3\to0$?
Write the closed ball of radius $R$ as an intersection of sets defined by bounded linear functionals. Why is each set in that intersection weakly closed?
What bounded functional shows that the coordinate vectors in $\ell^1$ do not converge weakly to zero?

Lower Semicontinuity of the Norm Under Weak Limits

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