Start With the Missing Hypothesis
A theorem can fail for many reasons, but a useful counterexample usually isolates one of them. Begin with a claim and identify the hypothesis whose removal is under examination. Then construct an object that satisfies the other hypotheses but violates the conclusion. If the example also violates a hypothesis that was supposed to remain in force, it does not diagnose the omission.
The previous tutorial repaired a claim about limits of differentiable functions by requiring uniform convergence of the derivatives and convergence at one anchor point. Those two conditions do different jobs: uniform control of the derivatives controls changes in the functions, while the anchor controls their values. We will test what can go wrong when either kind of control is absent. The aim is not merely to exhibit a failure, but to verify every retained hypothesis and locate the precise obstruction.
This is a logical discipline as well as a construction technique. For example, a sequence with derivatives that fail to converge uniformly says nothing about whether the anchor hypothesis is necessary if the sequence also fails to have convergent anchor values. Keep a short audit beside each proposed example: what is the domain, which assumptions hold, and exactly what conclusion fails?
A Moving-Point Test for Uniformity
Uniform convergence controls errors at all points of the domain at once. Thus, to disprove uniform convergence, it is often effective to choose a point \(x_n\) that depends on \(n\). At each fixed point, an error may tend to zero, while errors evaluated at carefully chosen moving points remain large.
Conversely, suppose \(f_n\) does not converge uniformly to \(f\). Negating the definition of uniform convergence, there is an \(\varepsilon_0>0\) such that for every positive integer \(N\), some \(n\geq N\) and some \(x\in E\) satisfy \(|f_n(x)-f(x)|\geq\varepsilon_0\). Choose such indices successively, with \(n_1<n_2<\cdots\), and points \(y_k\in E\) such that \(|f_{n_k}(y_k)-f(y_k)|\geq\varepsilon_0\). Define a sequence \((x_n)\) by setting \(x_{n_k}=y_k\) and choosing any point of \(E\) at other indices. If \(E\) is empty, uniform convergence holds vacuously, so in the nonuniform case these other choices are possible. The errors \(|f_{n_k}(x_{n_k})-f(x_{n_k})|\) are at least \(\varepsilon_0\) for every \(k\); therefore the errors along \((x_n)\) do not tend to zero. This proves the contrapositive and completes the proof. \(\square\)
In practice, the converse proof gives a recipe: if uniform convergence fails, there is a fixed positive error and a sequence of points witnessing it. A construction can run this logic in reverse by designing errors that are small at every fixed point but large at a moving point.
Worked Example: Pointwise Derivative Convergence with a Discontinuous Limit
For each positive integer \(n\), define \(f_n(x)=x^2/(x^2+1/n)\) on \(\mathbb R\). The denominator is positive for every real \(x\), so \(f_n\) is differentiable everywhere. Also, \(f_n(0)=0\) for every \(n\), giving a convergent anchor value.
At \(x=0\), \(f_n(x)=0\). At each fixed \(x\neq0\), multiplying numerator and denominator by \(n\) gives \(f_n(x)=nx^2/(nx^2+1)\), which tends to \(1\). Thus the pointwise limit is \(f(0)=0\) and \(f(x)=1\) for \(x\neq0\). This limit is discontinuous at zero: for \(x_k=1/k\), we have \(x_k\to0\), but \(f(x_k)=1\) for every \(k\).
The derivatives are \(f_n'(x)=(2x/n)/(x^2+1/n)^2=2nx/(nx^2+1)^2\). At \(x=0\), \(f_n'(0)=0\). For fixed \(x\neq0\), the last expression tends to zero, since its numerator has order \(n\) and its denominator has order \(n^2\). Therefore \(f_n'\) converges pointwise on \(\mathbb R\) to the zero function, even though the pointwise limit of \(f_n\) is discontinuous. The example satisfies differentiability, pointwise convergence of the functions, convergence of the anchor values, and pointwise convergence of the derivatives. It is uniform control of the derivatives that is missing.
The moving-point test makes that failure explicit. Set \(x_n=1/\sqrt n\). Then \(f_n'(x_n)=\sqrt n/2\), while the pointwise limit of the derivatives is zero everywhere. These errors do not tend to zero. In fact, they grow without bound, so the derivatives cannot converge uniformly to zero.
Omitting the Anchor
The derivative information in the repaired theorem controls differences between function values, not the absolute level of each function. Adding a constant to a function leaves its derivative unchanged. If those constants do not settle down, convergence of the derivatives cannot force the functions to converge.
Worked Example: Uniformly Convergent Derivatives but No Function Limit
Define \(f_n(x)=(-1)^n+x\) on \(\mathbb R\). Every \(f_n\) is differentiable, with \(f_n'(x)=1\) for every \(x\). Thus the derivatives converge uniformly to the constant function \(g(x)=1\): the error \(|f_n'(x)-g(x)|\) is exactly zero for every \(x\) and \(n\).
However, at the anchor \(x_0=0\), \(f_n(0)=(-1)^n\), which alternates between \(-1\) and \(1\) and does not converge. For any fixed \(x\), the sequence \(f_n(x)=x+(-1)^n\) also alternates between \(x-1\) and \(x+1\), so it does not converge. The conclusion that the functions have a pointwise limit fails, even though uniform convergence of the derivatives holds. The omitted anchor condition is exactly what would rule out this example.
This example also identifies a general construction move: when a condition concerns derivatives, try changing function values by constants. Such a change preserves every derivative while allowing the functions themselves to drift or oscillate.
Do Not Substitute a Weaker Kind of Convergence
A nearby but different mistake is to suppose that uniform convergence of the functions controls their derivatives. It does not. Derivatives measure local changes, and a small-amplitude function can still change rapidly. The next example verifies this distinction directly.
Worked Example: Uniformly Small Functions with Uncontrolled Derivatives
Let \(f_n(x)=\sin(nx)/n\) on \(\mathbb R\). Since \(|\sin(nx)|\leq1\), for every \(x\) we have \(|f_n(x)|\leq1/n\). Consequently, \(f_n\) converges uniformly to the zero function \(f(x)=0\).
But \(f_n'(x)=\cos(nx)\), whereas \(f'(x)=0\). At \(x=0\), \(f_n'(0)=1\) for every \(n\), so the derivatives do not even converge pointwise to \(f'(0)\). At \(x=\pi\), \(f_n'(\pi)=\cos(n\pi)=(-1)^n\), which does not converge at all. Thus uniform convergence of functions alone gives no conclusion that their derivatives converge to the derivative of the limit.
There is no contradiction with the result from the previous tutorial: that result requires uniform convergence of the derivatives, as well as convergence at an anchor. Here it is the functions that converge uniformly, while their derivatives do not satisfy the required control.
A Reliable Construction Procedure
The examples illustrate three different ways a conclusion can fail: a limit can be discontinuous when derivative convergence is only pointwise; functions can fail to converge when the anchor values do not converge; and derivatives can remain uncontrolled when only the functions converge uniformly. Before presenting any counterexample, check it in the same order as the theorem’s hypotheses.
Specify whether the issue is nonconvergence, discontinuity, nondifferentiability, or failure of uniform convergence.
For a derivative theorem, distinguish control of changes across the domain from control of the functions at one point.
Try a moving point when testing uniformity, or an additive constant when testing whether derivative information controls function values.
Verify the domain, differentiability, convergence type, and anchor values explicitly. Then show precisely why the conclusion fails.
A common pitfall is to treat pointwise and uniform convergence as interchangeable. Pointwise convergence fixes \(x\) first and then lets \(n\) grow; uniform convergence requires one index threshold to work for every \(x\). The moving-point criterion exposes the difference: uniform convergence controls errors even when the test point changes with \(n\). Another pitfall is to overlook additive constants. Derivatives cannot detect them, so a theorem that aims to control function values needs an anchor or another condition that serves the same purpose.
Check Your Understanding
Use the counterexample test and the examples above to answer these questions.
- In the moving-point criterion, how does the negation of uniform convergence produce a sequence of points with errors bounded below?
- For the quotient sequence, what is the pointwise limit at zero and at a nonzero point?
- Why does evaluating the quotient sequence’s derivative at \(x_n=1/\sqrt n\) rule out uniform convergence of the derivatives to zero?
- Which hypothesis fails for \(f_n(x)=(-1)^n+x\), despite uniform convergence of the derivatives?
- At which fixed point do the derivatives of \(\sin(nx)/n\) fail to converge to the derivative of the uniform limit?