Bounded Linear Operators Preserve Weak Convergence

Weak Convergence Through a Linear Map

Weak convergence is defined by testing vectors against continuous linear functionals. This definition makes one preservation principle almost inevitable: to test the images of a sequence under an operator, compose each target-space test with that operator. Boundedness is exactly what ensures that the resulting test is still continuous.

Let \(X\) and \(Y\) be Banach spaces over the same scalar field, either \(\mathbb{R}\) or \(\mathbb{C}\). Their continuous duals \(X^*\) and \(Y^*\) are the spaces of continuous linear functionals, equipped with the operator norm. A sequence \(x_n\) converges weakly to \(x\) in \(X\), written \(x_n\rightharpoonup x\), if \(\ell(x_n)\to\ell(x)\) for every \(\ell\in X^*\). These are scalar limits. A linear map \(T:X\to Y\) is bounded if there is a constant \(C<\infty\) such that \(\|Tx\|_Y\le C\|x\|_X\) for every \(x\in X\).

Theorem (Preservation of weak convergence): Let \(X,Y\) be Banach spaces and let \(T:X\to Y\) be a bounded linear map. If \(x_n\rightharpoonup x\) in \(X\), then \(Tx_n\rightharpoonup Tx\) in \(Y\).

Proof. To prove weak convergence in \(Y\), fix an arbitrary \(\lambda\in Y^*\). Define \(\ell:X\to\mathbb{K}\) by \(\ell(u)=\lambda(Tu)\), where \(\mathbb{K}\) is the scalar field. Linearity of \(T\) and \(\lambda\) gives linearity of \(\ell\), and boundedness gives, for every \(u\in X\),

|\ell(u)|=|\lambda(Tu)|\le \|\lambda\|_{Y^*}\|Tu\|_Y \le \|\lambda\|_{Y^*}\|T\|\|u\|_X.

Thus \(\ell\in X^*\). Since \(x_n\rightharpoonup x\), the definition of weak convergence gives \(\ell(x_n)\to\ell(x)\). Substituting the definition of \(\ell\), this is \(\lambda(Tx_n)\to\lambda(Tx)\). The functional \(\lambda\) was arbitrary in \(Y^*\), so \(Tx_n\rightharpoonup Tx\) in \(Y\). \(\square\)

Equivalently, each \(\lambda\in Y^*\) produces a functional \(T^*\lambda=\lambda\circ T\in X^*\), called the adjoint action of \(T\) on the duals. The estimate in the proof also shows \(\|T^*\lambda\|_{X^*}\le\|T\|\|\lambda\|_{Y^*}\). No compactness, injectivity, or surjectivity of \(T\) is required. Banach completeness is part of the stated setting, although the argument itself only needs normed spaces and continuous linear functionals.

Worked Example: Testing a Linear Map on a Weakly Convergent Sequence

In \(X=Y=\mathbb{R}^2\), use the Euclidean norm, and let \(x_n=(1,(-1)^n/n)\). Define \(T(a,b)=(a+2b,-a+b)\). Verify the weak limit of \(Tx_n\).

For an arbitrary functional \(\ell(a,b)=\alpha a+\beta b\), with \(\alpha,\beta\in\mathbb{R}\),

\ell(x_n)=\alpha+\beta\frac{(-1)^n}{n}\longrightarrow\alpha=\ell(1,0).

Every linear functional on \(\mathbb{R}^2\) is continuous, so this verifies \(x_n\rightharpoonup(1,0)\). Applying \(T\) gives

Tx_n=\left(1+2\frac{(-1)^n}{n},-1+\frac{(-1)^n}{n}\right)\longrightarrow(1,-1)=T(1,0).

For a direct functional check, take any \(\lambda(u,v)=\gamma u+\delta v\). Then \(\lambda(Tx_n)=\gamma(1+2(-1)^n/n)+\delta(-1+(-1)^n/n)\), which tends to \(\gamma-\delta=\lambda(1,-1)\). Check: when \(n=2\), \(x_2=(1,1/2)\) and \(Tx_2=(2,-1/2)\); the displayed formula gives those same coordinates. The limit also agrees with \(T(1,0)=(1,-1)\).

Inclusions Between \(L^p\) Spaces on Finite Measure Spaces

For a measure space \((X,\mathcal{M},\mu)\), \(L^p(\mu)\) denotes the space of equivalence classes, up to equality \(\mu\)-a.e., of measurable scalar-valued functions with finite \(p\)-norm; for \(p=\infty\), the norm is the essential supremum. We use the usual convention that \(1\le p\le\infty\). The following result concerns a finite measure space with \(0<\mu(X)<\infty\). No completeness assumption is needed. The case \(\mu(X)=0\) is degenerate: all these \(L^p\) spaces consist only of the zero equivalence class.

Theorem (Finite-measure inclusion): If \(1\le q\le p\le\infty\) and \(0<\mu(X)<\infty\), the identity on measurable functions defines a bounded linear inclusion \(I:L^p(\mu)\to L^q(\mu)\). For \(q<\infty\), if \(p<\infty\), then \(\|f\|_q\le\mu(X)^{1/q-1/p}\|f\|_p\); if \(p=\infty\), then \(\|f\|_q\le\mu(X)^{1/q}\|f\|_\infty\). For \(p=q=\infty\), \(\|f\|_\infty\le\|f\|_\infty\).

Proof. First suppose \(q<p<\infty\) and \(f\in L^p(\mu)\). Apply Hölder's inequality to \(|f|^q\) and the constant function \(1\), with conjugate exponents \(r=p/q>1\) and \(r'=p/(p-q)>1\). The second function has \(L^{r'}\) norm \(\mu(X)^{1/r'}\), which is finite. Therefore

\int_X |f|^q\,d\mu \le \left(\int_X |f|^{qr}\,d\mu\right)^{1/r} \left(\int_X 1^{r'}\,d\mu\right)^{1/r'} =\|f\|_p^q\mu(X)^{1-q/p}.

Taking the \(q\)-th root gives the asserted bound. If \(q=p<\infty\), the bound is equality because the exponent of \(\mu(X)\) is zero. If \(q<\infty\) and \(p=\infty\), then \(|f|\le\|f\|_\infty\) \(\mu\)-a.e., so integrating \(|f|^q\le\|f\|_\infty^q\) gives the stated estimate. When \(p=q=\infty\), the map is the identity and its norm is \(1\). These estimates show both that every \(L^p\) equivalence class belongs to \(L^q\) and that the identity map is bounded. It is well-defined on equivalence classes because equality \(\mu\)-a.e. is the same equivalence relation in the two spaces. Linearity follows from the identity on representatives. \(\square\)

Hölder's inequality is used here in its standard form: for conjugate exponents \(r,r'>1\), \(\int|uv|\,d\mu\le\|u\|_r\|v\|_{r'}\). The inclusion result has a direct weak-convergence consequence.

Corollary: Under the finite-measure assumptions above, if \(f_n\rightharpoonup f\) in \(L^p(\mu)\) and \(1\le q\le p\le\infty\), then the same equivalence classes satisfy \(f_n\rightharpoonup f\) in \(L^q(\mu)\).

Proof. The inclusion \(I:L^p(\mu)\to L^q(\mu)\) is bounded and linear by the theorem. Apply preservation of weak convergence to \(I\): \(If_n\rightharpoonup If\) in \(L^q(\mu)\). Since \(I\) sends each function class to that same class, this is the asserted conclusion. \(\square\)

Worked Example: Shrinking Intervals in \(L^4\) and \(L^2\)

On \([0,1]\) with Lebesgue measure \(m\), let \(f_n=\chi_{(0,1/n)}\). Show that \(f_n\rightharpoonup0\) in \(L^4(m)\), and use the inclusion to obtain weak convergence in \(L^2(m)\).

For \(g\in L^{4/3}(m)\), Hölder's inequality gives

\left|\int_0^1 f_ng\,dm\right| =\left|\int_{(0,1/n)}g\,dm\right| \le \|g\|_{4/3}\,m((0,1/n))^{1/4} =\|g\|_{4/3}\,n^{-1/4}\longrightarrow0.

The standard duality theorem for \(L^p\) on this finite Lebesgue measure space says that every continuous linear functional on \(L^4(m)\) is represented by integration against a function in \(L^{4/3}(m)\) (with the usual conjugation convention in the complex case). Thus the calculation tests every functional in \(L^4(m)^*\), and \(f_n\rightharpoonup0\) in \(L^4(m)\). Since \(0< m([0,1])=1<\infty\) and \(2\le4\), the inclusion theorem gives weak convergence in \(L^2(m)\) as well. The norms can also be checked directly: \(m((0,1/n))=1/n\), so \(\|f_n\|_4=(1/n)^{1/4}=n^{-1/4}\) and \(\|f_n\|_2=(1/n)^{1/2}=n^{-1/2}\). Both tend to zero. Check: for \(n=16\), the interval has measure \(1/16\), so the norms are \(16^{-1/4}=1/2\) and \(16^{-1/2}=1/4\), respectively.

Worked Example: A Sharp Inclusion Bound and a Failed Reverse Inclusion

On \([0,4]\) with Lebesgue measure, the inclusion \(L^2(m)\to L^1(m)\) has bound \(\|f\|_1\le2\|f\|_2\), since \(m([0,4])^{1-1/2}=4^{1/2}=2\). For \(f=\chi_{[0,4]}\), \(\|f\|_1=\int_0^4 1\,dm=4\) and \(\|f\|_2=(\int_0^4 1\,dm)^{1/2}=2\). Thus \(4=2\cdot2\), so this bound is attained.

Reversing the exponents does not give an inclusion \(L^1(m)\to L^2(m)\). For \(f(x)=x^{-3/4}\) on \(0<x\le4\), with any finite value assigned at \(x=0\),

\int_0^4 |f(x)|\,dx =\int_0^4 x^{-3/4}\,dx =\left[4x^{1/4}\right]_0^4 =4\sqrt{2}<\infty,

whereas \(\int_0^4|f(x)|^2\,dx=\int_0^4x^{-3/2}\,dx=\infty\), since the antiderivative \(-2x^{-1/2}\) diverges at \(0\). Hence \(f\in L^1(m)\) but \(f\notin L^2(m)\). Check: the integrability distinction comes from the powers: \(x^{-3/4}\) has finite integral at zero, while \(x^{-3/2}\) does not. Therefore there is no identity inclusion from all of \(L^1(m)\) into \(L^2(m)\), even though the interval has finite measure.

What the Hypotheses Do—and Do Not—Give

A common mistake is to assume that finite measure makes the \(L^p\) inclusions go in both directions. The estimate proved above instead runs from the larger exponent to the smaller one: when \(q\le p\), \(L^p(\mu)\subseteq L^q(\mu)\). The function in the preceding example is a concrete correction to the false reverse inclusion. It also shows why the order of exponents is not a cosmetic detail.

Finite measure matters too. On \(\mathbb{R}\) with Lebesgue measure, the indicators \(u_N=\chi_{(0,N)}\) satisfy \(\|u_N\|_2=N^{1/2}\) and \(\|u_N\|_1=N\). Their ratio \(\|u_N\|_1/\|u_N\|_2=N^{1/2}\) is unbounded as \(N\to\infty\). Thus no constant \(C\) can make \(\|u\|_1\le C\|u\|_2\) for every such indicator, and there is no bounded identity inclusion \(L^2(\mathbb{R})\to L^1(\mathbb{R})\). Each \(u_N\) belongs to both spaces; the calculation specifically rules out a uniform operator bound.

The operator theorem itself is more general than the \(L^p\) application. Whenever a sequence converges weakly in a space and a bounded linear map is available, composing target functionals gives weak convergence of the images. But one must first verify that the proposed map is defined on the whole source space and is bounded. An inclusion that is not defined for every source element, or an identity map with no uniform norm bound, cannot be used in the theorem.

Check Your Understanding

Answer each question by identifying the functional test or the norm estimate that applies.

In the preservation theorem, why does \(\lambda\circ T\) belong to \(X^*\) when \(\lambda\in Y^*\)? State the norm bound.
For a finite measure space with \(0<\mu(X)<\infty\), which direction of inclusion holds when \(1\le q\le p\le\infty\)? Give the bound when \(q< p<\infty\).
Why does weak convergence in \(L^4([0,1])\) imply weak convergence in \(L^2([0,1])\)? Name the bounded map used.
Does finite measure imply \(L^1(\mu)\subseteq L^2(\mu)\)? Use \(x^{-3/4}\) on \([0,4]\) to justify your answer.
On \(\mathbb{R}\) with Lebesgue measure, use \(\chi_{(0,N)}\) to test whether the identity \(L^2\to L^1\) has a uniform bound.