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Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.

For the free Schrodinger equation with compatible smooth initial and boundary data, subtract $g(t)e^{-x}$ before taking a Fourier sine transform. The lifted initial datum vanishes at the endpoint, making its sine transform $O(k^{-2})$ if its second derivative is integrable. The transformed forcing is $k(ig-g^{\prime })/(1+k^2)$; time integration followed by integration by parts makes its solution contribution uniformly $O(k^{-3})$. An integrable spectral majorant proves uniform convergence even at the compatible initial-boundary corner.

If real polynomials $p_n$ converge by uniform convergence to $f$ on all of the real line, then for some $N$ and constants $c_n$, $p_n=p_N+c_n$ for every $n\ge N$, with $c_n$ convergent. Indeed, a common uniform error below one gives $|p_n-p_N|<2$ everywhere. A bounded polynomial on the real line is constant, since every nonconstant polynomial grows in absolute value at infinity. Evaluating at zero identifies the limit of a sequence of constants, so $f=p_N+c$ is a polynomial. This conclusion depends on the entire real line being the domain.

A uniform limit of continuous functions is a continuous function. No compactness assumption on the domain of a function is required.^[ref]

For a proof, fix an input $x$ and tolerance $\epsilon >0$. Uniform convergence supplies one $f_N$ with $\Vert f_N-f{\Vert }_{\infty }<\epsilon /3$. Its continuity at $x$ supplies $\delta >0$ with $|f_N(y)-f_N(x)|<\epsilon /3$ for allowed inputs satisfying $|y-x|<\delta$. The triangle inequality givesOne approximating function controls both endpoints of this estimate. Under pointwise convergence, selecting an index for $x$ supplies no common control at nearby $y$. This explains why the polynomial functions $x^n$ can have a discontinuous pointwise limit on $[0,1]$.

For each positive integer $n$, put $f_n(t)=\sin (2\pi nt)/n$ on $[0,1]$. Its supremum norm is exactly $1/n$, so this sequence of functions converges to zero by uniform convergence. Nevertheless,Each of its $n$ complete sine oscillations contributes $4/n$: rise from zero to $1/n$, fall through zero to $-1/n$, then return to zero. Taking a partition of an interval through all extrema attains $4$. Every further partition refinement merely subdivides monotone function segments, so the same reasoning as in partition example of total variation proves the upper bound too.

For a continuously differentiable function on a closed bounded interval, total variation also satisfiesApplied here, the derivative is $2\pi \cos (2\pi nt)$, and integrating its absolute value over $n$ periods gives $4$. The differentiability hypothesis matters: the partition definition applies much more widely than this derivative formula.^[ref] Small pointwise size and small accumulated movement are different approximation goals.

On the half-line, $\sup f_n=1/(en)\to 0$, but ${\int }_0^{\infty }{f}_n=1$. The change of variable $x=ny$ shows why: the decreasing amplitude is spread over an increasing interval. Uniform convergence controls integrals on finite-measure domains; it does not alone control total mass on an infinite-measure domain.

The supremum of pointwise distances defines a metric on maps when it is finite, for example if the target metric space is bounded. Convergence in this metric is uniform convergence. If the target is complete, Cauchy sequences have pointwise limits and the Cauchy estimate makes convergence uniform.

The Poisson kernel on the circle is nonnegative, has normalized integral one, and its mass outside any fixed arc about zero tends to zero as $r\uparrow 1$. Split the convolution error into a short arc, controlled by the modulus of continuity, and its complement, controlled by the vanishing kernel mass. This proves uniform convergence for every continuous periodic function. Its Fourier multiplier is $r^{|n|}$, so the convolution is the Abel sum of the Fourier series.

On $[0,1]$, the polynomials $p_n(x)=x^n/n$ converge in supremum norm to the zero function, since $\Vert p_n{\Vert }_{\infty }=1/n$. But $p_n^{\prime }(1)=1$ for every $n$, whereas the zero function has left derivative zero at that endpoint. Thus uniform convergence of function values does not imply uniform convergence of derivatives. This does not contradict uniform convergence of Bernstein polynomial derivatives, which has a particular approximation construction and additional differentiability hypotheses.