Repair a False Real Analysis Theorem

A Plausible Claim That Fails

A useful way to test a proposed theorem is to ask what its hypotheses actually control. Pointwise convergence controls the values of a sequence at each fixed input, but it does not control how rapidly those values change near that input. Differentiability depends on precisely this local change. Consequently, the claim that pointwise limits of differentiable functions must be differentiable is false.

False Claim: If a sequence of differentiable functions converges pointwise, then its limit is differentiable.

To repair the claim, we will identify a stronger form of control that does suffice. Uniform convergence of the derivatives limits the change in each function across an interval, while convergence of the functions at one fixed point prevents the whole sequence from drifting by an arbitrary additive constant. The Mean Value Theorem turns these two pieces of information into a differentiable limit.

Worked Example: Smooth Functions with a Nondifferentiable Limit

For each positive integer $n$, define $f_n(x)=\sqrt{x^2+1/n}$ on $\mathbb R$. The quantity under the square root is strictly positive, so each $f_n$ is differentiable, with $f_n'(x)=x/\sqrt{x^2+1/n}$. For every fixed $x$, $x^2+1/n$ tends to $x^2$, and therefore $f_n(x)$ tends to $f(x)=|x|$.

The limit is not differentiable at zero: for $h>0$, $(|h|-|0|)/h=1$, whereas for $h<0$, $(|h|-|0|)/h=-1$. Thus the two-sided derivative does not exist. This directly disproves the false claim.

The derivatives also show why the example evades the repair we will prove. At each fixed $x>0$, $f_n'(x)$ tends to $1$; at each fixed $x<0$, it tends to $-1$; and at zero it is always zero. But this convergence is not uniform. At $x_n=1/\sqrt n$, $f_n'(x_n)=1/\sqrt2$, while the pointwise limit of the derivatives at $x_n$ is $1$. The difference is $1-1/\sqrt2$, independent of $n$, so it cannot be made uniformly small.

The Estimate That Guides the Repair

The relevant estimate compares two functions by comparing their derivatives and their values at one point. Let $I$ be an interval, let $x_0\in I$, and suppose $u$ and $v$ are continuous on $I$ and differentiable on its interior. Apply the Mean Value Theorem to $u-v$ on the segment joining $x_0$ to $x$. When $x\ne x_0$, this gives $|(u-v)(x)-(u-v)(x_0)|\leq |x-x_0|\sup_{t\in I^\circ}|u'(t)-v'(t)|$. For $x=x_0$, the left side is zero, so the same estimate holds.

Lemma (Anchored Mean Value Estimate): Let $I$ be an interval with nonempty interior, and let $u,v:I\to\mathbb R$ be continuous on $I$ and differentiable on $I^\circ$. For any $x_0,x\in I$, $$ |(u-v)(x)-(u-v)(x_0)|\leq |x-x_0|\sup_{t\in I^\circ}|u'(t)-v'(t)|. $$ If the supremum is infinite, the inequality is understood as a trivial extended bound.

Proof: If $x=x_0$, both sides are zero or nonnegative, so the inequality holds. If $x\ne x_0$, the closed segment with endpoints $x_0$ and $x$ lies in $I$. The function $u-v$ is continuous on this segment and differentiable at every point in its interior. By the Mean Value Theorem, for some point $c$ strictly between $x_0$ and $x$, $(u-v)(x)-(u-v)(x_0)=(x-x_0)(u'(c)-v'(c))$. Taking absolute values and bounding the derivative difference by its supremum proves the estimate. $\square$

This estimate explains why the value at $x_0$ matters. If the derivatives of two functions are close, their difference changes little as we move away from $x_0$. But their difference at $x_0$ could still be large. Convergence at the anchor point controls that remaining freedom.

A Corrected Theorem

Theorem (Uniform Convergence of Derivatives): Let $I\subseteq\mathbb R$ be an interval with nonempty interior, and fix $x_0\in I$. For each positive integer $n$, let $f_n:I\to\mathbb R$ be continuous on $I$ and differentiable on $I^\circ$. Suppose $f_n(x_0)$ converges to a real number and $f_n'$ converges uniformly on $I^\circ$ to a function $g:I^\circ\to\mathbb R$. Then $f_n$ converges pointwise on $I$ to a function $f$, and $f$ is differentiable on $I^\circ$ with $f'(x)=g(x)$ for every $x\in I^\circ$.

The interval need not be bounded. Pointwise convergence of $f_n$ follows from the anchored estimate, applied to pairs of terms. Differentiability requires a second use of the same control: increments of the limit can be compared to increments of one fixed $f_n$, whose derivative at the point is close to $g$.

Proof. Uniform convergence of $f_n'$ implies that $(f_n')$ is uniformly Cauchy on $I^\circ$. Indeed, given $\varepsilon>0$, choose $N$ such that $|f_n'(t)-g(t)|<\varepsilon/2$ for all $n\geq N$ and all $t\in I^\circ$. Then, for $n,m\geq N$, $|f_n'(t)-f_m'(t)|<\varepsilon$ for all such $t$. Also, the convergent sequence $(f_n(x_0))$ is Cauchy.

Fix $x\in I$. Applying the Anchored Mean Value Estimate to $f_n$ and $f_m$ gives $|f_n(x)-f_m(x)|\leq |f_n(x_0)-f_m(x_0)|+|x-x_0|\sup_{t\in I^\circ}|f_n'(t)-f_m'(t)|$. For this fixed $x$, both terms on the right tend to zero as $n,m$ tend to infinity. Thus $(f_n(x))$ is Cauchy in $\mathbb R$, so it has a real limit. Define $f(x)=\lim_{n\to\infty}f_n(x)$ for each $x\in I$.

It remains to prove the derivative assertion. Fix $x\in I^\circ$ and $\varepsilon>0$. By uniform convergence, choose $N$ such that $\sup_{t\in I^\circ}|f_k'(t)-g(t)|<\varepsilon/8$ for all $k\geq N$, and fix any $n\geq N$. Then $\sup_{t\in I^\circ}|f_n'(t)-g(t)|<\varepsilon/4$. For every $m\geq n$, the triangle inequality gives $\sup_{t\in I^\circ}|f_m'(t)-f_n'(t)|<\varepsilon/2$. If $h\ne0$ is small enough that $x+h\in I$, the Mean Value Theorem applied to $f_m-f_n$ on the segment from $x$ to $x+h$ yields $|(f_m(x+h)-f_m(x))-(f_n(x+h)-f_n(x))|\leq (\varepsilon/2)|h|$. Letting $m$ tend to infinity is valid because $f_m$ converges pointwise at both $x$ and $x+h$. Hence $|(f(x+h)-f(x))-(f_n(x+h)-f_n(x))|\leq(\varepsilon/2)|h|$.

Since $f_n$ is differentiable at $x$, there is a $\delta>0$ such that, whenever $0<|h|<\delta$ and $x+h\in I$, $\left|(f_n(x+h)-f_n(x))/h-f_n'(x)\right|<\varepsilon/4$. For these $h$, divide the preceding increment estimate by $|h|$, then use the triangle inequality and $|f_n'(x)-g(x)|<\varepsilon/4$. The result is $\left|(f(x+h)-f(x))/h-g(x)\right|<\varepsilon/2+\varepsilon/4+\varepsilon/4=\varepsilon$. This is the definition of $f'(x)=g(x)$. Since $x$ was arbitrary in $I^\circ$, the proof is complete. $\square$

Using the Repair

Worked Example: A Small Perturbation of a Quadratic

Define $f_n(x)=x^2+\sin(x)/n$ on $\mathbb R$. Each function is differentiable, and $f_n(0)=0$ for all $n$, so the anchor values converge. Pointwise, $f_n(x)$ tends to $f(x)=x^2$. Moreover, $f_n'(x)=2x+\cos(x)/n$, and $\sup_{x\in\mathbb R}|f_n'(x)-2x|\leq 1/n$. Thus the derivatives converge uniformly on $\mathbb R$ to $g(x)=2x$. The theorem applies and concludes that the limit is differentiable with derivative $2x$, in agreement with direct differentiation of $x^2$.

The example also illustrates why the theorem can apply on an unbounded interval: the derivatives $f_n'$ themselves are unbounded, but their differences from the limit derivative are uniformly bounded by $1/n$. Uniform convergence concerns the error, not the size of each function.

Worked Example: Geometric Polynomial Partial Sums

Fix $r$ with $0<r<1$, and on $I=[0,r]$ define $f_n(x)=\sum_{k=1}^{n}x^k$. The finite geometric-sum formula gives $f_n(x)=x(1-x^n)/(1-x)$, so $f_n(x)$ tends to $f(x)=x/(1-x)$ on this interval. Also $f_n(0)=0$.

Differentiating the finite sum gives $f_n'(x)=\sum_{k=1}^{n}k x^{k-1}$. For $x\in[0,r]$, the tail of the corresponding series is bounded by $\sum_{k=n+1}^{\infty}k r^{k-1}=r^n\left((n+1)/(1-r)+r/(1-r)^2\right)$, which tends to zero. Thus the derivatives converge uniformly. The finite-sum identity $\sum_{k=1}^{n}k x^{k-1}=[1-(n+1)x^n+nx^{n+1}]/(1-x)^2$ shows that their limit is $g(x)=1/(1-x)^2$: the terms involving $n r^n$ tend to zero uniformly for $x\in[0,r]$. The theorem therefore gives $f'(x)=1/(1-x)^2$ on the interior of $[0,r]$, matching the derivative of $x/(1-x)$.

Worked Example: Why the Anchor Value Cannot Be Omitted

Let $f_n(x)=n+x$ on $\mathbb R$. Every derivative is the constant function $f_n'(x)=1$, so the derivatives converge uniformly to $g(x)=1$. Nevertheless, the functions do not converge pointwise to a real-valued function: for each fixed $x$, $n+x$ tends to infinity. The missing hypothesis is convergence at the anchor point; for example, $f_n(0)=n$ does not converge in $\mathbb R$.

This example is not a counterexample to the theorem. It shows that uniform derivative convergence controls changes in the functions but does not control their absolute levels. A single convergent value supplies that missing control.

What the Repair Does—and Does Not—Say

The corrected theorem is a package of two distinct requirements. Uniform convergence of the derivatives is stronger than pointwise convergence: it keeps the derivative error small at every point of the interval at once. Convergence at one point anchors the functions so that the derivative control can be converted into control of their values. With both hypotheses, the proof first obtains pointwise convergence and then verifies the derivative of the limit directly from difference quotients.

A common pitfall is to prove only that the derivatives converge pointwise and then use the theorem anyway. The first example shows why this fails: points close to zero can have derivative errors that stay large even though the derivatives converge at every fixed point. Another pitfall is to assume uniform convergence of the functions alone repairs differentiability. The false claim at the start required no such conclusion; the relevant added control here is on the derivatives, together with an anchor value.

Check the proposed conclusion.
For a differentiability claim, examine whether the hypotheses control local changes, not just pointwise values.

Locate the missing control.
Pointwise derivative convergence may leave large errors at moving points; uniform convergence rules out that failure.

Anchor the sequence.
Require convergence at one point so derivative estimates also control the functions themselves.

Prove the repaired conclusion.
Use the Mean Value Theorem for function increments, then pass to the limit in a difference quotient.

Check Your Understanding

Use the counterexample, estimate, and theorem to answer these questions.

At what point does the pointwise limit in the first worked example fail to be differentiable?
Why does evaluating the derivative error at $x_n=1/\sqrt n$ disprove uniform convergence in that example?
What does the anchored Mean Value Estimate control, and what does it leave uncontrolled without an anchor value?
In the theorem’s proof, why is it valid to let $m$ tend to infinity in the estimate for the increments of $f_m-f_n$?
Which hypothesis fails for the sequence $f_n(x)=n+x$, even though its derivatives converge uniformly?