Uniform Limits of Continuous Functions

What Uniform Limits Preserve

A sequence of continuous functions can approach its limit at every input without approaching it uniformly. That distinction matters: pointwise convergence alone does not guarantee that the limit is continuous. Uniform convergence gives stronger control, and the earlier theorem Uniform Limits of Continuous Functions Are Continuous establishes that a uniform limit of continuous functions is continuous. Here we use that result to examine what uniform approximation preserves and how to recognize when uniform convergence is impossible.

The same estimates that make a limit continuous can also preserve more specific features, provided those features have a quantitative form shared by every function in the sequence. A common Lipschitz bound is one example. Convexity is another: it is expressed by an inequality that can be passed to a pointwise, and therefore to a uniform, limit.

Definition: A function \(f:E\to\mathbb{R}\) is Lipschitz with constant \(L\geq0\) if \[ |f(x)-f(y)|\leq L|x-y| \] for every \(x,y\in E\). A function \(f:I\to\mathbb{R}\), where \(I\) is an interval, is convex if \[ f(tx+(1-t)y)\leq t f(x)+(1-t)f(y) \] for every \(x,y\in I\) and \(t\in[0,1]\).

A Lipschitz condition gives one bound on changes in function values across the entire domain. Convexity compares a function at an intermediate point with a weighted average of its endpoint values. Both properties are useful because they can be stated as inequalities and checked directly for a limit.

A Shared Lipschitz Bound Passes to the Limit

Suppose every function in a sequence satisfies the same Lipschitz estimate. If the functions converge uniformly, then at any two inputs their values converge to the corresponding limit values. Passing to the limit in the shared estimate shows that the limit satisfies it too.

Theorem (Uniform Limits Preserve a Common Lipschitz Bound): Let \(E\subseteq\mathbb{R}\), and suppose \(f_n:E\to\mathbb{R}\) is Lipschitz with the same constant \(L\geq0\) for every \(n\). If \(f_n\) converges uniformly on \(E\) to \(f\), then \(f\) is Lipschitz with constant \(L\).

Proof. Fix \(x,y\in E\). Uniform convergence implies pointwise convergence, so \(f_n(x)\to f(x)\) and \(f_n(y)\to f(y)\). For every \(n\), the common Lipschitz bound gives

|f_n(x)-f_n(y)|\leq L|x-y|.

Taking the limit as \(n\to\infty\), continuity of absolute value yields

|f(x)-f(y)|\leq L|x-y|.

Since \(x\) and \(y\) were arbitrary, this inequality holds for every pair of inputs in \(E\). Thus \(f\) is Lipschitz with constant \(L\). \(\square\)

The uniform convergence hypothesis is stronger than the pointwise convergence used in this short proof, but it is often the way the limit is known to have been constructed. The shared constant is essential to the conclusion as stated: knowing only that each \(f_n\) is Lipschitz, with a bound that may grow without limit, supplies no single estimate to pass to the limit.

Worked Example: Approximating Absolute Value with a Shared Lipschitz Bound

On \([-1,1]\), define \[ f_n(x)=\left(1-\frac{1}{n}\right)|x|+\frac{x}{n}, \qquad n\geq1. \] Each \(f_n\) is continuous. For any \(x,y\in[-1,1]\), the triangle inequality and \(||x|-|y||\leq|x-y|\) give

|f_n(x)-f_n(y)| \leq \left(1-\frac{1}{n}\right)||x|-|y||+\frac{1}{n}|x-y| \leq |x-y|.

Thus every \(f_n\) has the same Lipschitz constant \(1\). To check uniform convergence to \(f(x)=|x|\), compute

|f_n(x)-|x|| =\frac{1}{n}\bigl|x-|x|\bigr| \leq\frac{2}{n},

because \(|x-|x||=0\) for \(x\geq0\) and \(|x-|x||=2|x|\leq2\) for \(-1\leq x<0\). The bound \(2/n\) tends to zero independently of \(x\), so convergence is uniform. The theorem confirms that the limit \(|x|\) retains the common Lipschitz bound \(1\), including at the point \(x=0\), where the formula for the limit has a corner.

Uniform Limits Also Preserve Convexity

Convexity is expressed by an inequality involving three function values. For fixed \(x,y\), and \(t\), each value in that inequality converges along the sequence. Passing to the limit therefore preserves the inequality.

Theorem (Uniform Limits Preserve Convexity): Let \(I\) be an interval. Suppose each \(f_n:I\to\mathbb{R}\) is convex and \(f_n\) converges uniformly on \(I\) to \(f\). Then \(f\) is convex on \(I\).

Proof. Fix \(x,y\in I\) and \(t\in[0,1]\). Since \(I\) is an interval, \(z=tx+(1-t)y\) belongs to \(I\). Convexity of \(f_n\) gives

f_n(z)\leq t f_n(x)+(1-t)f_n(y)

for every \(n\). Uniform convergence implies convergence at each of the three fixed inputs \(z,x,y\). Taking limits in the inequality gives

f(z)\leq t f(x)+(1-t)f(y).

This is the defining inequality for convexity. Since the choices of \(x,y\), and \(t\) were arbitrary, \(f\) is convex on \(I\). \(\square\)

Worked Example: A Uniform Limit of Convex Functions

On \([-2,2]\), let \(g_n(x)=x^2+1/n\). Each function is continuous. To verify convexity directly, take \(x,y\in[-2,2]\) and \(t\in[0,1]\). The constant terms cancel in the convexity comparison, and

t g_n(x)+(1-t)g_n(y)-g_n(tx+(1-t)y) =t(1-t)(x-y)^2\geq0.

Thus every \(g_n\) is convex. Also,

|g_n(x)-x^2|=\frac{1}{n}

for every \(x\in[-2,2]\). Since this error is independent of \(x\) and tends to zero, \(g_n\) converges uniformly to \(g(x)=x^2\). The convexity theorem ensures that the limit remains convex; the defining inequality can also be checked directly from the displayed calculation.

These results illustrate a useful distinction. Continuity is guaranteed for a uniform limit of continuous functions by the theorem established earlier in this course. Other properties may also pass to the limit, but one should identify the specific inequality or condition that supports the conclusion. A common Lipschitz constant and the convexity inequality each provide such a condition.

When Uniform Approximation Is Impossible

The continuity theorem also supplies a practical obstruction. If a discontinuous function were a uniform limit of continuous functions, it would have to be continuous. Therefore, no sequence of continuous functions can converge uniformly to a discontinuous target. The next example makes that obstruction quantitative for a jump.

Worked Example: Continuous Approximations to a Jump Are Not Uniform

Define \(H:[-1,1]\to\mathbb{R}\) by \(H(x)=0\) for \(x<0\), \(H(0)=1/2\), and \(H(x)=1\) for \(x>0\). For each \(n\geq1\), define

q_n(x)=\frac{1}{2}\left(1+\frac{x}{\sqrt{x^2+1/n}}\right).

The denominator is positive for every \(x\), so \(q_n\) is continuous on \([-1,1]\). At \(x=0\), \(q_n(0)=1/2=H(0)\). For fixed \(x>0\),

\frac{x}{\sqrt{x^2+1/n}}\longrightarrow 1, \qquad\text{so}\qquad q_n(x)\longrightarrow1=H(x).

For fixed \(x<0\), the same ratio tends to \(-1\), so \(q_n(x)\to0=H(x)\). Thus \(q_n\) converges pointwise to \(H\). But at the moving input \(x_n=1/\sqrt n\),

q_n(x_n) =\frac{1}{2}\left(1+\frac{1/\sqrt n}{\sqrt{1/n+1/n}}\right) =\frac{1}{2}\left(1+\frac{1}{\sqrt2}\right), \qquad H(x_n)=1.

Consequently,

|q_n(x_n)-H(x_n)|=\frac{1}{2}\left(1-\frac{1}{\sqrt2}\right)>0

for every \(n\). The errors therefore cannot tend uniformly to zero. More generally, every continuous function \(g\) on \([-1,1]\) has uniform error at least \(1/2\) when approximating \(H\). If \(M=\sup_{x\in[-1,1]}|g(x)-H(x)|\), continuity at zero and limits from the left and right give \(|g(0)|\leq M\) and \(|g(0)-1|\leq M\). Hence

1\leq |g(0)|+|g(0)-1|\leq 2M, \qquad\text{so}\qquad M\geq\frac12.

The example shows both how pointwise convergence can hide a moving region of error and why continuity prevents that error from disappearing uniformly across a jump.

How to Use the Continuity Theorem Carefully

When a sequence of continuous functions is given, first identify its candidate limit and then check whether convergence is uniform on the domain in question. If it is, the earlier theorem gives continuity of the limit without requiring a separate continuity argument at every point. If convergence is only pointwise, that theorem does not apply, and the limit may be discontinuous, as the jump example demonstrates.

The domain matters. A sequence may converge uniformly on a smaller set but not on a larger one, so a conclusion based on uniform convergence is valid only on the domain where the uniform estimate holds. Likewise, a shared Lipschitz constant must be independent of both the index and the inputs, and convexity must hold for every function in the sequence.

Takeaway: Uniform convergence preserves continuity for a sequence of continuous functions. It can also preserve shared quantitative structure, such as a common Lipschitz bound, and inequality-defined structure, such as convexity. Pointwise convergence alone does not provide these conclusions.

Check Your Understanding

Use the estimates and results in this tutorial to answer the following questions.

Why does the proof that a common Lipschitz bound passes to the limit fix two inputs before taking limits?
Which part of the convexity proof uses the assumption that the domain is an interval?
For the functions \(f_n(x)=(1-1/n)|x|+x/n\), what uniform error bound proves convergence to \(|x|\) on \([-1,1]\)?
At which moving input does the sequence \(q_n\) retain a fixed positive error from \(H\)?
Why can the continuity theorem rule out uniform convergence to a jump, even when pointwise convergence occurs?