Approximation Counterexamples

When Approximation Must Fail

The applications of Stone-Weierstrass show how a unital algebra that separates points can approximate every continuous function on a compact set. It is just as useful to recognize situations in which approximation cannot work. A proposed approximation family may obey a symmetry, a fixed-value condition, or a boundedness restriction that the target does not share. Such constraints can give an explicit positive lower bound on the best possible error.

Throughout the compact-set examples, let $K$ be a nonempty compact subset of $\mathbb{R}$, and use the supremum norm on $C(K)$. A lower bound matters: it proves not merely that a particular sequence of approximants fails, but that every element of the proposed family stays a definite distance from the target. We begin with a general test based on bounded linear functionals.

Theorem (Annihilating Functional Error Bound): Let $X$ be a normed space, let $A\subseteq X$, and let $L:X\to\mathbb{R}$ be a bounded linear functional such that $L(g)=0$ for every $g\in A$. If $L$ is not the zero functional, then for every $f\in X$, $$ \inf_{g\in A}\|f-g\|\geq\frac{|L(f)|}{\|L\|}. $$

Proof. Fix any $g\in A$. Since $L(g)=0$, linearity gives $L(f)=L(f-g)$. The definition of the operator norm implies

|L(f)|=|L(f-g)|\leq\|L\|\|f-g\|.

Because $L$ is nonzero, $\|L\|>0$, so division by $\|L\|$ gives $\|f-g\|\geq |L(f)|/\|L\|$. This inequality holds for every $g\in A$. Taking the infimum over $A$ proves the result. $\square$

In $C(K)$, point evaluation is bounded: for $a\in K$, $|f(a)|\leq\|f\|_\infty$. Differences of point evaluations are bounded as well. Thus, if every candidate approximant takes equal values at two points but the target does not, the difference of those evaluations gives an immediate error bound. The argument is not limited to algebras; it applies to any family contained in the kernel of a bounded linear functional.

Worked Examples: Symmetry and Fixed-Value Obstructions

Worked Example: Even Polynomials Cannot Approximate the Identity

Let $A$ be the set of even polynomials restricted to $[-1,1]$, and take $f(x)=x$. Every $g\in A$ satisfies $g(1)=g(-1)$. Define $L:C[-1,1]\to\mathbb{R}$ by $L(h)=h(1)-h(-1)$. This is linear and bounded, because

|L(h)|\leq |h(1)|+|h(-1)|\leq2\|h\|_\infty.

Its norm is exactly $2$: the bound above gives $\|L\|\leq2$, and for the constant function $h=1$, $\|h\|_\infty=1$ and $L(h)=0$, which does not show equality. Instead take $h(x)=x$, for which $\|h\|_\infty=1$ and $L(h)=2$; hence $\|L\|\geq2$. Therefore $\|L\|=2$. The functional vanishes on $A$, while $L(f)=1-(-1)=2$. The theorem gives

\inf_{g\in A}\|x-g\|_\infty\geq\frac{2}{2}=1.

The zero polynomial attains error $\|x\|_\infty=1$, so the best error is exactly $1$. The obstruction is symmetry: an even function cannot match the target's different values at $1$ and $-1$. In particular, the fact that polynomials are dense in $C[-1,1]$ does not mean that every restricted family of polynomials is dense.

Worked Example: Polynomials with Zero Constant Term Cannot Approximate One

On $[0,1]$, consider $A=\{p:p\text{ is a polynomial and }p(0)=0\}$. Every element of $A$ vanishes at $0$. Let $L(h)=h(0)$. Since $|L(h)|\leq\|h\|_\infty$, and $L(1)=1=\|1\|_\infty$, its norm is $1$. It vanishes on $A$, whereas $L(1)=1$. Consequently,

\inf_{p\in A}\|1-p\|_\infty\geq1.

This bound is exact: choosing $p=0$ gives $\|1-0\|_\infty=1$. Thus this particular nonunital polynomial family is not dense in $C[0,1]$. This example does not say that every nonunital algebra fails to be dense; the earlier result on constants in the uniform closure shows why that broader conclusion would be unwarranted. The point is to inspect the actual constraint: here every candidate is forced to take the value zero at the left endpoint.

The two examples use the same test in different forms. The first compares two points that the approximants cannot distinguish; the second uses a single point at which every approximant has a prescribed value. More generally, whenever a bounded linear functional annihilates the candidate family but not the target, the theorem supplies a quantitative obstruction.

Why the Whole Real Line Changes the Problem

The compactness of the domain in polynomial approximation results is not a technical detail that can simply be dropped. On a compact interval, every polynomial is bounded, and the supremum norm measures its error over a finite domain. On the whole real line, a nonconstant polynomial is unbounded. Therefore it cannot be within finite supremum distance of a bounded function.

Proposition: Every real polynomial that is bounded on $\mathbb{R}$ is constant. Consequently, if $f\in C_b(\mathbb{R})$ is not constant, then no sequence of polynomials can converge uniformly to $f$ on $\mathbb{R}$.

Proof. Suppose $p$ has degree $n\geq1$ and leading coefficient $a_n\ne0$. Write

p(x)=a_nx^n+\sum_{j=0}^{n-1}a_jx^j.

For $x\ne0$, divide by $x^n$. As $x\to+\infty$, each term $a_j/x^{n-j}$ tends to zero, so $p(x)/x^n\to a_n$. Since $a_n\ne0$, there is an $R>0$ such that $|p(x)/x^n|\geq |a_n|/2$ for $x\geq R$. Hence $|p(x)|\geq (|a_n|/2)x^n$ there, and $p$ is unbounded. Thus a bounded polynomial must have degree zero and be constant.

Now suppose polynomials $p_m$ converge uniformly on $\mathbb{R}$ to a bounded function $f$. For some index $m$, $\|p_m-f\|_{\infty,\mathbb{R}}<\infty$. Since $f$ is bounded, the triangle inequality gives $|p_m(x)|\leq\|p_m-f\|_{\infty,\mathbb{R}}+\|f\|_{\infty,\mathbb{R}}$ for every $x$. Thus $p_m$ is bounded, hence constant. Every later polynomial $p_k$ is also at finite uniform distance from $f$, and therefore bounded and constant. A uniform limit of constant functions is constant: if $p_k(x)=c_k$, then $|f(x)-f(y)|\leq|f(x)-c_k|+|c_k-f(y)|\to0$ for any $x,y$. So $f$ must be constant, proving the assertion. $\square$

Worked Example: Polynomials Cannot Uniformly Approximate Sine on the Real Line

The function $f(x)=\sin x$ is bounded and nonconstant on $\mathbb{R}$. By the proposition, no sequence of polynomials can converge uniformly to it on the whole real line. There is also a direct error bound. If a polynomial $p$ has finite uniform distance from $\sin x$, then $p$ is bounded, so it is a constant $c$. Since sine takes both values $1$ and $-1$, its error satisfies

\|\sin-c\|_{\infty,\mathbb{R}} \geq\max\{|1-c|,|-1-c|\}\geq1.

For the last inequality, the triangle inequality gives $2=|1-(-1)|\leq|1-c|+|c-(-1)|$, so at least one of the two terms is at least $1$. Taking $c=0$ gives error exactly $1$, so the best possible error among polynomials at finite uniform distance is $1$. This does not conflict with polynomial approximation on compact intervals: the domain and the norm have changed.

Pointwise Approximation Is Not Uniform Approximation

A different pitfall is to treat pointwise convergence as though it were uniform. Pointwise convergence controls the error at each fixed point, with the allowed index depending on that point. Uniform convergence requires one index to control the error at every point simultaneously. The distinction is essential because uniform limits preserve continuity, while pointwise limits need not.

Proposition (Continuity of Uniform Limits): Let $K\subseteq\mathbb{R}$, and suppose each $f_n:K\to\mathbb{R}$ is continuous. If $f_n$ converges uniformly to $f$ on $K$, then $f$ is continuous on $K$.

Proof. Fix $x_0\in K$ and $\varepsilon>0$. Uniform convergence gives an index $N$ such that $|f_N(x)-f(x)|<\varepsilon/3$ for every $x\in K$. By continuity of $f_N$ at $x_0$, there is a $\delta>0$ such that, for $x\in K$ with $|x-x_0|<\delta$, we have $|f_N(x)-f_N(x_0)|<\varepsilon/3$. The triangle inequality then yields

|f(x)-f(x_0)| \leq|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)| <\varepsilon.

Thus $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous on $K$. $\square$

Worked Example: A Pointwise Polynomial Limit That Is Not Uniform

On $[0,1]$, set $f_n(x)=x^n$. Each $f_n$ is a polynomial and hence continuous. If $0\leq x<1$, then $x^n\to0$; at $x=1$, $x^n=1$ for every $n$. Thus the pointwise limit is $f(1)=1$ and $f(x)=0$ for $0\leq x<1$, which is discontinuous at $1$.

The convergence cannot be uniform: the proposition would force the limit to be continuous. Directly, for every $n$, the error at $x=1$ is zero, but for points $x<1$ arbitrarily close to $1$, the error is $x^n$, which can be made arbitrarily close to $1$. Therefore

\sup_{0\leq x\leq1}|f_n(x)-f(x)|=1 \qquad\text{for every }n.

The supremum is $1$ even though it is not attained on $[0,1)$: values $x^n$ approach $1$ as $x$ approaches $1$ from below. Pointwise approximation alone therefore gives no uniform approximation conclusion.

Using Counterexamples as a Diagnostic

When an approximation claim seems plausible, first identify the domain and the norm. Then inspect what every candidate must satisfy. Does the family preserve a symmetry? Does it impose a fixed value? Is a bounded global norm being used on a noncompact domain? Is the claimed convergence only pointwise? Each question can reveal a barrier before any complicated approximation argument is attempted.

The annihilating-functional theorem is especially useful because it turns a structural restriction into a numerical lower bound. Evaluation differences detect forced equalities at two points; a point evaluation detects forced values at one point. On the other hand, the noncompact example shows that some obstructions are even more basic: the candidate functions may not lie at finite distance from the target at all. These arguments do not replace density theorems. They clarify which hypotheses and which notion of approximation make a density conclusion possible.

Takeaway: To disprove density, find a constraint shared by every candidate but violated by the target. A bounded linear functional that vanishes on the approximation family converts that constraint into an explicit lower bound on the uniform error.

Check Your Understanding

Use the obstruction tests and examples above to answer the following questions.

If a bounded linear functional $L$ vanishes on every candidate approximant, what lower bound does it give for approximating $f$, provided $L(f)\ne0$?
Why does the equality $g(1)=g(-1)$ for every even polynomial prevent uniform approximation of $x$ on $[-1,1]$ with error below $1$?
Which bounded linear functional detects the obstruction for polynomials on $[0,1]$ that all vanish at $0$?
Why can a nonconstant bounded function on $\mathbb{R}$ not be a uniform limit of polynomials?
What continuity property rules out uniform convergence of $x^n$ to its pointwise limit on $[0,1]$?

What You'll Learn