Continuous Mapping Problems for Norms and Maxima

From Vector Limits to Limits of Summaries

A random vector often matters because of a numerical summary derived from it. Its length, its largest coordinate, or a polynomial expression in its coordinates may be more relevant than the vector itself. Weak convergence provides a systematic way to pass from a vector limit to the limit of such summaries.

The main task is to identify the function being applied and check its continuity. Norms, coordinate maxima, and polynomials are continuous on finite-dimensional Euclidean spaces. Consequently, once the random vectors converge jointly, the continuous mapping theorem gives the weak limit of each transformed vector.

Key principle: If \(X_n\) converges weakly to \(X\) in \(\mathbb{R}^d\), and \(f:\mathbb{R}^d\to\mathbb{R}^k\) is continuous, then \(f(X_n)\) converges weakly to \(f(X)\). The limit is obtained by applying the same map to the limiting random vector.

Here and below, \(X_n\Rightarrow X\) denotes weak convergence. The important input is convergence of the full vector, not merely convergence of each coordinate separately.

The Maps of Interest

For \(x=(x_1,\ldots,x_d)\in\mathbb{R}^d\), common summaries include a norm \(\lVert x\rVert\), the largest coordinate \(\max_{1\le i\le d}x_i\), and the largest absolute coordinate \(\max_{1\le i\le d}|x_i|\). These last two summaries are different: negative coordinates can affect the maximum absolute value without affecting the coordinate maximum.

Definition: A norm on \(\mathbb{R}^d\) is a function \(\lVert\cdot\rVert:\mathbb{R}^d\to[0,\infty)\) such that \(\lVert x\rVert=0\) exactly when \(x=0\), \(\lVert ax\rVert=|a|\lVert x\rVert\) for every scalar \(a\), and \(\lVert x+y\rVert\le \lVert x\rVert+\lVert y\rVert\) for all \(x,y\). The Euclidean norm is \(\lVert x\rVert_2=(\sum_{i=1}^d x_i^2)^{1/2}\); another common choice is \(\lVert x\rVert_\infty=\max_i|x_i|\).

A polynomial map is a function with one or more polynomial coordinate functions. For example, \(f:\mathbb{R}^2\to\mathbb{R}^2\) given by \(f(x,y)=(x^2+xy,y^3)\) is a polynomial map. The output can be scalar-valued or vector-valued.

Result: Every norm, coordinate maximum, maximum absolute coordinate, and polynomial map on a finite-dimensional Euclidean space is continuous. Thus each is eligible for the continuous mapping theorem.

Why Norms and Maxima Are Continuous

The triangle inequality gives a useful quantitative bound for any norm. For maxima, changing every coordinate by a small amount can change the maximum by no more than the largest coordinate change. These bounds establish continuity directly.

Proposition: For any norm on \(\mathbb{R}^d\) and all \(x,y\in\mathbb{R}^d\), \[ |\lVert x\rVert-\lVert y\rVert|\le \lVert x-y\rVert. \] In particular, a norm is continuous.

Proof. The triangle inequality applied to \(x=(x-y)+y\) gives \(\lVert x\rVert\le \lVert x-y\rVert+\lVert y\rVert\), so \(\lVert x\rVert-\lVert y\rVert\le\lVert x-y\rVert\). Interchanging \(x\) and \(y\) gives \(\lVert y\rVert-\lVert x\rVert\le\lVert y-x\rVert=\lVert x-y\rVert\). Together these inequalities yield the claimed absolute-value bound. As \(y\) tends to \(x\) in the norm, the bound forces \(\lVert y\rVert\) to tend to \(\lVert x\rVert\). In finite dimensions every norm induces the usual Euclidean topology, so this is continuity on \(\mathbb{R}^d\). \(\square\)

Proposition: Define \(m(x)=\max_{1\le i\le d}x_i\) and \(a(x)=\max_{1\le i\le d}|x_i|\). Then \[ |m(x)-m(y)|\le \max_i|x_i-y_i| \quad\text{and}\quad |a(x)-a(y)|\le \max_i|x_i-y_i|. \] Both maps are continuous.

Proof. Let \(\delta=\max_i|x_i-y_i|\). For every index \(i\), \(x_i\le y_i+\delta\le m(y)+\delta\), so \(m(x)\le m(y)+\delta\). Reversing \(x\) and \(y\) gives \(m(y)\le m(x)+\delta\), proving the first bound. Also, \(||x_i|-|y_i||\le |x_i-y_i|\le\delta\). Therefore \(|x_i|\le |y_i|+\delta\le a(y)+\delta\) for every \(i\), which gives \(a(x)\le a(y)+\delta\). Reversing the roles again yields the second bound. Each bound implies continuity. \(\square\)

Polynomial maps are continuous because coordinate functions, sums, and products of real-valued continuous functions are continuous. The coordinate functions \(x\mapsto x_i\) are continuous, and each polynomial is built from them using finitely many sums and products.

Applying the Continuous Mapping Theorem

The continuous mapping theorem states that if \(X_n\Rightarrow X\) in \(\mathbb{R}^d\) and \(f:\mathbb{R}^d\to\mathbb{R}^k\) is continuous, then \(f(X_n)\Rightarrow f(X)\). For the functions above, continuity holds everywhere, so there are no exceptional points to check under the law of \(X\).

Practical procedure: First specify the vector \(X_n\) and its joint weak limit \(X\). Next write the desired statistic as a function \(f\) of the whole vector. Verify that \(f\) is continuous, then identify the limiting distribution as the law of \(f(X)\).

The theorem determines the limit in distribution, not generally a numerical distribution formula. To find such a formula, evaluate \(f(X)\) using the known law of \(X\). In particular, dependence among the coordinates of \(X\) is automatically retained by applying \(f\) to the joint vector.

Worked Examples

Worked Example: A Norm with a Two-Point Limit

Let \(X\) equal \((3,4)\) with probability \(1/3\) and \((0,0)\) with probability \(2/3\). Set \(X_n=X+(1/n,0)\). For any bounded continuous \(h:\mathbb{R}^2\to\mathbb{R}\), the expectation \(\mathbb{E}[h(X_n)]\) is the weighted sum of \(h(3+1/n,4)\) and \(h(1/n,0)\). Continuity of \(h\) shows that this sum tends to \(\mathbb{E}[h(X)]\); hence \(X_n\Rightarrow X\).

Take \(f(x)=\lVert x\rVert_2\), which is continuous. The continuous mapping theorem gives \(\lVert X_n\rVert_2\Rightarrow\lVert X\rVert_2\). The limiting norm is \(5\) when \(X=(3,4)\), since \(\sqrt{3^2+4^2}=5\), and is \(0\) when \(X=(0,0)\). Thus the limiting distribution puts probability \(1/3\) at \(5\) and probability \(2/3\) at \(0\).

The perturbed norms also show these values directly: they are \(\sqrt{(3+1/n)^2+16}\) and \(1/n\), respectively, and converge to \(5\) and \(0\).

Worked Example: Coordinate and Absolute Maxima

Let \(X=(2,-1)\) with probability \(0.4\) and \(X=(-3,1)\) with probability \(0.6\), and again set \(X_n=X+(0,1/n)\). As in the preceding finite-support argument, bounded continuous test functions show \(X_n\Rightarrow X\).

For the coordinate maximum \(m(x)=\max(x_1,x_2)\), the two limiting values are \(m(2,-1)=2\) and \(m(-3,1)=1\). Therefore \(m(X_n)\Rightarrow m(X)\), whose distribution assigns probability \(0.4\) to \(2\) and \(0.6\) to \(1\).

For the maximum absolute coordinate \(a(x)=\max(|x_1|,|x_2|)\), the values are \(a(2,-1)=2\) and \(a(-3,1)=3\). Hence \(a(X_n)\Rightarrow a(X)\), with probability \(0.4\) at \(2\) and \(0.6\) at \(3\). The two transformed limits differ because the negative coordinate \(-3\) does not determine the coordinate maximum, but it does determine the absolute maximum.

Worked Example: A Polynomial of a Random Vector

Let \(X=(A,B)\) equal \((1,2)\) with probability \(1/3\) and \((-1,1)\) with probability \(2/3\). Define \(X_n=(A+1/n,B-1/n)\). For every bounded continuous \(h\), the expectation of \(h(X_n)\) is a sum of two terms with fixed weights, and each term converges to the corresponding value of \(h(X)\). Thus \(X_n\Rightarrow X\).

Consider the polynomial \(q(x,y)=x^2+xy+2y^2\). Its values at the two possible limit vectors are \[ q(1,2)=1^2+(1)(2)+2(2^2)=1+2+8=11, \] and \[ q(-1,1)=(-1)^2+(-1)(1)+2(1^2)=1-1+2=2. \] Since \(q\) is continuous, \(q(X_n)\Rightarrow q(X)\). The limit puts probability \(1/3\) at \(11\) and \(2/3\) at \(2\).

Using Several Summaries at Once

Several continuous summaries can be combined into one vector-valued map. For example, a statistic might record a norm, a maximum coordinate, and a polynomial of the same random vector. Because each component is continuous, the combined map is continuous as well.

Result: Suppose \(X_n\Rightarrow X\) in \(\mathbb{R}^d\), and let \(f_1,\ldots,f_r\) be continuous maps from \(\mathbb{R}^d\) to Euclidean spaces. Then \[ \bigl(f_1(X_n),\ldots,f_r(X_n)\bigr) \Rightarrow \bigl(f_1(X),\ldots,f_r(X)\bigr). \]

Proof. Define \(F(x)=(f_1(x),\ldots,f_r(x))\), with the outputs placed in the product Euclidean space. If \(x_n\to x\), continuity of each \(f_j\) gives \(f_j(x_n)\to f_j(x)\) for every \(j\). Since there are finitely many components, these convergences imply \(F(x_n)\to F(x)\), so \(F\) is continuous. The continuous mapping theorem applied to \(F\) now gives \(F(X_n)\Rightarrow F(X)\), which is the stated convergence. \(\square\)

Common Pitfalls and Why Joint Convergence Matters

A transformation such as \(x_1x_2\) or \(\max(x_1,x_2)\) depends on how the coordinates occur together. Separate convergence of the coordinate distributions does not, by itself, establish the joint weak convergence needed to apply the theorem to such a function. The input should be a verified vector limit \(X_n\Rightarrow X\), not just a list of marginal limits.

Another common error is to transform the marginal limits independently and then combine the results as if the limiting coordinates were independent. The continuous mapping theorem makes no independence assumption: the joint law of \(X\) determines the law of \(f(X)\). A further distinction worth keeping explicit is whether a problem asks for \(\max_i X_{n,i}\), \(\max_i|X_{n,i}|\), or a norm; they are different maps and can have different limits.

Once the map is identified, the calculation is often short. The substantive checks are that the vector converges jointly, the specified map is continuous, and the final distribution is computed under the actual joint law of the limit.

Check Your Understanding

Use the continuous mapping theorem and the continuity bounds above to answer these questions.

If \(X_n\Rightarrow X\) in \(\mathbb{R}^d\), what is the weak limit of \(\lVert X_n\rVert_\infty\)?
For \(x=(-4,2)\), what are \(\max_i x_i\) and \(\max_i|x_i|\)? Why are they different?
Why does continuity of a polynomial map allow it to be applied to a weakly convergent random vector?
What joint-distribution information is needed to find the law of \(X_1X_2\)?
If \(f_1\) and \(f_2\) are continuous and \(X_n\Rightarrow X\), what is the joint weak limit of \((f_1(X_n),f_2(X_n))\)?