When Differentiation and Limits Commute

Passing Differentiation Through a Limit

Uniform convergence of functions controls their values, but differentiation depends on how those values change over arbitrarily short distances. The previous tutorial showed that uniform convergence of functions alone does not guarantee that their limit is differentiable. To justify passing a limit through differentiation, we need control of the derivatives as well as information fixing the functions’ values.

The Uniform Derivative Convergence Criterion, established earlier in this course, gives a sufficient condition: uniform convergence of the derivatives, together with convergence of the functions at one point, lets us identify the derivative of the limit. Here we consider a useful local version. It applies even when derivative convergence is not uniform across an entire interval, provided it is uniform on every compact subinterval.

Definition: Let $I$ be an open interval. A sequence $g_n:I\to\mathbb{R}$ converges locally uniformly to $g:I\to\mathbb{R}$ if it converges uniformly on every compact subinterval of $I$. Equivalently, for every compact subinterval $K\subset I$, $\sup_{x\in K}|g_n(x)-g(x)|\to0$.

Local uniform convergence allows the uniform error bound to depend on the compact interval being considered. This distinction matters on unbounded intervals: control on each bounded region need not give one error bound valid everywhere. An anchor point supplies the remaining information needed to determine the values of the limit function.

A Local Criterion for Commuting Limits and Derivatives

Theorem (Local Uniform Derivative Criterion): Let $I$ be an open interval, and suppose each $f_n:I\to\mathbb{R}$ is continuously differentiable. Assume that $f_n'$ converges locally uniformly on $I$ to $g$, and that $f_n(a)$ converges to a finite real number for some $a\in I$. Then $f_n$ converges locally uniformly on $I$ to a function $f$, and $f$ is differentiable with $f'=g$ on $I$.

Proof. Choose any compact subinterval $K\subset I$. There is a compact subinterval $J\subset I$ whose interior contains both $K$ and $a$. By local uniform convergence, $f_n'$ converges uniformly on $J$. The anchor values $f_n(a)$ converge by hypothesis. Apply the Uniform Derivative Convergence Criterion to the functions restricted to the interior of $J$. It follows that $f_n$ converges uniformly there to a differentiable function whose derivative is $g$.

In particular, the sequence converges uniformly on $K$, and its limit is differentiable on the interior of $J$. If two such intervals $J$ overlap, their resulting limits agree on the overlap: at each point there, both are limits of the same real sequence $f_n(x)$, and a sequence of real numbers has at most one limit. Thus the local limits define one function $f$ on all of $I$. Since $K$ was arbitrary, convergence to $f$ is locally uniform. Since every point of $I$ lies in the interior of such an interval $J$, $f$ is differentiable throughout $I$, with $f'=g$. $\square$

The anchor condition is not an unnecessary extra assumption when the theorem is used to deduce convergence of the functions. Derivatives determine changes in function values, but do not determine a function’s overall vertical position. Once the value at one point is controlled, the derivative information determines the limiting values elsewhere.

Worked Example: Local Convergence Without Global Uniform Convergence

For $n\geq1$, define $f_n(x)=x/n$ on $\mathbb{R}$. Then $f_n'(x)=1/n$, so the derivatives converge uniformly on $\mathbb{R}$, and hence locally uniformly, to $g(x)=0$. Also $f_n(0)=0$ for every $n$. The theorem gives the limit $f(x)=0$, with $f'(x)=0=g(x)$.

The function convergence is uniform on each compact interval: if $K\subset[-R,R]$, then $$ \sup_{x\in K}|f_n(x)-0|\leq \frac{R}{n}\longrightarrow0. $$ But the convergence is not uniform on all of $\mathbb{R}$. For each $n$, the errors $|x|/n$ are unbounded as $x$ ranges over $\mathbb{R}$. Thus even global uniform convergence of the derivatives does not in general imply global uniform convergence of the functions on an unbounded interval. The local conclusion is the appropriate one here.

An Error Bound for Changes in Function Values

The criterion also yields a useful quantitative statement. If an approximating derivative is close to its limit on a particular interval, then the change in the error $f-f_N$ between two points of that interval is small. This controls increments rather than absolute function values, so it does not depend on a new choice of anchor.

Theorem (Increment Error Estimate): Under the hypotheses of the Local Uniform Derivative Criterion, fix $N\geq1$ and $x,y\in I$. Then $$ \left|(f(y)-f(x))-(f_N(y)-f_N(x))\right| \leq |y-x|\sup_{t\in[x,y]}|g(t)-f_N'(t)|. $$ Here $[x,y]$ denotes the closed interval with endpoints $x$ and $y$.

Proof. The claim is immediate if $x=y$, since both sides are zero. Otherwise, the interval with endpoints $x$ and $y$ is compact and lies inside $I$. For $m\geq1$, the function $f_m-f_N$ is continuous on that interval and differentiable in its interior. The Mean Value Theorem gives $$ |(f_m(y)-f_N(y))-(f_m(x)-f_N(x))| \leq |y-x|\sup_{t\in[x,y]}|f_m'(t)-f_N'(t)|. $$ As $m\to\infty$, local uniform convergence of $f_m'$ to $g$ implies that the supremum on the right converges to $\sup_{t\in[x,y]}|g(t)-f_N'(t)|$. Indeed, the functions $f_m'-f_N'$ converge uniformly on $[x,y]$ to $g-f_N'$, and uniform convergence implies convergence of their supremum norms by the reverse triangle inequality for the supremum norm.

The left side converges to the left side of the claimed estimate because the Local Uniform Derivative Criterion gives $f_m(x)\to f(x)$ and $f_m(y)\to f(y)$. Taking limits in the inequality proves the result. $\square$

This estimate can be used when an explicit approximation $f_N$ is easier to work with than the limit $f$. It bounds the error in the change from $x$ to $y$ using only the derivative error on the segment between them. It does not, by itself, bound $|f(x)-f_N(x)|$ at an individual point: for that, one also needs information at an anchor point.

Worked Example: Using the Increment Error Estimate

Return to $f_n(x)=x/n$ on $\mathbb{R}$, with $f(x)=0$ and $g(x)=0$. For any fixed $N$, the derivative error is constant: $$ \sup_{t\in[x,y]}|g(t)-f_N'(t)|=\frac1N. $$ The Increment Error Estimate therefore gives $$ \left|(f(y)-f(x))-(f_N(y)-f_N(x))\right| \leq \frac{|y-x|}{N}. $$ The left side can be calculated directly: $$ \left|0-\left(\frac{y}{N}-\frac{x}{N}\right)\right| =\frac{|y-x|}{N}. $$ Thus equality holds. The estimate records precisely how the derivative error controls the error in increments, even though the function errors are not uniformly bounded on $\mathbb{R}$.

Why the Anchor and Uniformity Matter

The next example shows why convergence at an anchor cannot be dropped when the theorem is used to establish convergence of the functions. The derivatives can be perfectly controlled while the functions drift vertically without bound.

Worked Example: Derivatives Do Not Fix Vertical Position

Let $f_n(x)=n+x$ on $\mathbb{R}$. Then $f_n'(x)=1$ for every $n$ and $x$, so the derivatives converge uniformly to the constant function $1$. But $f_n(0)=n$ does not converge to a finite real number. In fact, for each fixed $x$, $f_n(x)=n+x\to+\infty$, so there is no finite pointwise limit function on $\mathbb{R}$.

The derivatives describe the shared slope but reveal nothing about the increasing constants $n$. This is exactly the information the anchor condition supplies: it prevents the whole sequence from escaping vertically.

It is important to distinguish two uses of convergence at an anchor. If pointwise convergence of $f_n$ to a finite function $f$ is already known on $I$, then $f_n(a)\to f(a)$ automatically for every $a\in I$. Pointwise convergence of the functions therefore does satisfy the anchor condition. It does not, however, imply uniform convergence of the derivatives. Conversely, pointwise convergence of the derivatives alone does not replace their uniform convergence when one wants to pass differentiation through the limit.

Worked Example: Pointwise Derivative Limits Can Give the Wrong Value

For $n\geq1$, define $$ f_n(x)=x+\frac{x}{1+nx^2} \qquad (-1\leq x\leq1). $$ For $t=|x|$, the added term has magnitude $t/(1+nt^2)$. Its maximum for $t\geq0$ occurs at $t=1/\sqrt{n}$, where it equals $1/(2\sqrt{n})$. Hence $$ \sup_{x\in[-1,1]}|f_n(x)-x| \leq\frac1{2\sqrt{n}}\longrightarrow0, $$ so $f_n$ converges uniformly, and therefore pointwise, to $f(x)=x$.

Differentiating gives $$ f_n'(x)=1+\frac{1-nx^2}{(1+nx^2)^2}. $$ At the origin, $f_n'(0)=2$ for every $n$, whereas $f'(0)=1$. For any fixed $x\neq0$, the fraction tends to zero, since its numerator has order $n$ and its denominator has order $n^2$. Thus $f_n'(x)\to1=f'(x)$ for every fixed nonzero $x$, but $f_n'(0)$ does not converge to $f'(0)$. In particular, the derivative convergence is not uniform: the error at $x=0$ is always $1$.

This example demonstrates a specific limitation, not a failure of the criterion: the derivatives do not converge pointwise to $f'$ at every point, and they certainly do not converge uniformly. Pointwise convergence of the functions, even uniform convergence in this example, cannot make up for the missing control on the derivatives.

Interpreting the Commutation Rule

The conclusion $f'=g$ says that, under the stated hypotheses, differentiating the limiting function gives the same result as taking the limit of the derivatives. Local uniform convergence is enough because differentiation is a local operation: to determine $f'(x)$, it suffices to control the sequence on a compact interval around $x$. The anchor condition is global in the sense that it fixes the level of the functions, while local uniform convergence controls their changes.

The Uniform Derivative Convergence Criterion provides the underlying principle on a single interval with uniform derivative control. The local version applies that principle on compact subintervals and ensures that the resulting limits agree wherever those subintervals overlap. The Increment Error Estimate adds a practical bound: on any fixed segment, the derivative approximation controls how accurately the approximating function reproduces the limiting function’s change.

Takeaway: To pass differentiation through a sequence limit locally, control the derivatives uniformly on compact subintervals and ensure convergence at one anchor point. Pointwise convergence of derivatives alone is not a substitute for that uniform control.

Check Your Understanding

Use the criterion, estimate, and examples in this tutorial to answer the following questions.

What does locally uniform convergence of the derivatives require on each compact subinterval?
Why does convergence of the derivatives alone fail to determine the limiting function in the example $f_n(x)=n+x$?
If $f_n$ is already known to converge pointwise to a finite function on $I$, does the sequence of values at a chosen anchor point converge?
In the increment error estimate, why is the supremum taken over the segment joining $x$ and $y$?
For the final worked example, what are $f_n'(0)$ and $f'(0)$, and what do they show about pointwise derivative convergence?