Pointwise limits do not preserve boundedness or sup norms (Tag PR-LCV2V)
For \(n\geq1\), the functions \(f_n\colon(0,1)\to\RR\), \(f_n(x)\coloneqq\min(n,1/x)\), are bounded and converge pointwise to the unbounded function \(x\mapsto1/x\).
For \(n\geq1\), the functions \(g_n\coloneqq\chi_{[n,n+1]}\colon\RR\to\RR\) converge pointwise to \(0\), while \[ \lim_{n\to\infty}\sup_{x\in\RR}\abs{g_n(x)}=1\neq0=\sup_{x\in\RR}\abs{\lim_{n\to\infty}g_n(x)} . \]
A sequence of bounded functions with unbounded pointwise limit (Tag FE-5JE3Z)
A pointwise limit of bounded functions need not be bounded. For \(n\geq 1\) let \(f_n\colon(0,1)\to\RR\), \(f_n(x) \coloneqq \frac{1}{x + \frac1n}\). Each \(f_n\) is bounded, since \(0 < f_n(x) \leq n\) for \(x\in(0,1)\). For each \(x\in(0,1)\), \(f_n(x)\to \frac1x\), and \(x\mapsto\frac1x\) is unbounded on \((0,1)\).
Pointwise limits of continuous functions need not be continuous (Tag PR-SCTER)
For \(n\geq1\) let \(f_n\colon[0,1]\to\RR\), \(f_n(x)\coloneqq x^n\). Each \(f_n\) is continuous, and \(f_n\) converges pointwise to the discontinuous function \(f\) with \(f(x)=0\) for \(0\leq x<1\) and \(f(1)=1\). For \(x_k\coloneqq1-1/k\), the iterated limits differ: \[ \lim_{k\to \infty} \lim_{n\to\infty} f_n(x_k) = 0 \neq 1 = \lim_{n\to \infty} \lim_{k\to\infty} f_n(x_k) . \]
$x^n$ converges pointwise but not uniformly on $[0,1]$ (Tag FE-MDTXW)
For \(n\geq 1\) let \(f_n\colon[0,1]\to\RR\), \(f_n(x)\coloneqq x^n\). Then \(f_n\to f\coloneqq\chi_{\theset{1}}\) pointwise on \([0,1]\), since \(x^n\to 0\) for \(0\leq x<1\) and \(1^n = 1\). The convergence is not uniform: for every \(n\), \(\sup_{x\in[0,1]}\abs{f_n(x) - f(x)} = \sup_{0\leq x<1} x^n = 1\).
Limits of differentiable functions need not be differentiable (Tag PR-DWOXP)
There are differentiable functions \(f_n\colon\RR\to\RR\) that converge uniformly on \(\RR\) to a function that is not differentiable.
There are differentiable functions \(f_n\colon\RR\to\RR\) that converge uniformly on \(\RR\) to a differentiable function \(f\) while \(f_n'(0)\not\to f'(0)\).
Let \(f_n(x)\coloneqq\sqrt{x^2+1/n}\). Each \(f_n\) is differentiable, and \[ 0\le f_n(x)-\abs{x}=\frac{1/n}{\sqrt{x^2+1/n}+\abs{x}}\le\frac{1}{\sqrt n}, \] so \(f_n\to\abs{x}\) uniformly on \(\RR\), and \(\abs{x}\) is not differentiable at \(0\).
Let \(f_n(x)\coloneqq n^{-1/2}\sin(nx)\). Then \(\sup_x\abs{f_n(x)}\le n^{-1/2}\), so \(f_n\to f=0\) uniformly, but \(f_n'(0)=n^{1/2}\to\infty\neq0=f'(0)\).
If \(f_n\colon[a,b]\to\RR\) are differentiable, \((f_n(x_0))\) converges for some \(x_0\in[a,b]\), and \((f_n')\) converges uniformly on \([a,b]\), then \((f_n)\) converges uniformly on \([a,b]\) to a differentiable function \(f\) with \(f'=\lim_n f_n'\). A uniform limit of differentiable functions can be differentiable at no point: by the Weierstrass approximation theorem, every continuous function on \([a,b]\), including a nowhere differentiable one, is a uniform limit of polynomials.
A uniformly convergent sequence of differentiable functions whose derivatives do not converge pointwise (Tag FE-73BWZ)
For \(n\geq 1\) let \(f_n\colon\RR\to\RR\), \(f_n(x) \coloneqq \frac{\sin(nx)}{n}\). Then \(\abs{f_n(x)}\leq\frac1n\) for all \(x\), so \(f_n\to 0\) uniformly on \(\RR\). Each \(f_n\) is differentiable with \(f_n'(x) = \cos(nx)\), and \(f_n'(\pi) = \cos(n\pi) = (-1)^n\), so the sequence \((f_n'(\pi))\) diverges and \((f_n')\) does not converge pointwise on \(\RR\).
Differentiable functions converging to $\abs{x}$ (Tag FE-GN3CF)
A limit of differentiable functions need not be differentiable. For \(n\geq 1\) let \(f_n\colon\RR\to\RR\), \(f_n(x) \coloneqq \frac{x^2}{\sqrt{x^2 + \frac1n}}\). Each \(f_n\) is differentiable on \(\RR\), since \(x^2+\frac1n > 0\). For every \(x\in\RR\), \[ \abs{\abs{x} - f_n(x)} = \frac{\abs{x}}{\sqrt{x^2+\frac1n}}\qty{\sqrt{x^2+\tfrac1n} - \abs{x}} \leq \frac{1/n}{\sqrt{x^2+\frac1n} + \abs{x}} \leq \frac{1}{\sqrt n}, \] so \(f_n\to\abs{x}\) uniformly, and in particular pointwise, on \(\RR\). The limit \(x\mapsto\abs{x}\) is not differentiable at \(0\).
Uniformly convergent differentiable $f_n\to f$ with $f_n'\to g$ pointwise and $g\neq f'$ (Tag FE-2UZDB)
A sequence of differentiable functions can converge uniformly to a differentiable function \(f\) while the derivatives converge pointwise to a function \(g\neq f'\).
For \(n\geq 1\) let \(f_n\colon\RR\to\RR\), \(f_n(x) \coloneqq \frac{x}{1+nx^2}\). Since \(1+nx^2\geq 2\sqrt{n}\abs{x}\), we have \(\abs{f_n(x)}\leq \frac{1}{2\sqrt n}\) for all \(x\in\RR\), with equality at \(x = 1/\sqrt n\). Hence \(\norm{f_n}_\infty = \frac{1}{2\sqrt n}\to 0\), and \(f_n\to f\coloneqq 0\) uniformly on \(\RR\).
The derivatives are \[ f_n'(x) = \frac{1-nx^2}{(1+nx^2)^2}. \] For every \(n\), \(f_n'(0) = 1\); for \(x\neq 0\), \(\abs{f_n'(x)}\leq \frac{1+nx^2}{(1+nx^2)^2} = \frac{1}{1+nx^2}\to 0\). Hence \(f_n'\to g\coloneqq\chi_{\theset{0}}\) pointwise on \(\RR\), while \(f' = 0\) and \(g(0) = 1\neq f'(0)\).
Limits and integrals need not commute (Tag PR-UZVC3)
For \(n\geq1\) let \(f_n\coloneqq n\chi_{(0,1/n)}\colon[0,1]\to\RR\). Then \(f_n(x)\to0\) for every \(x\in[0,1]\), but \[ \lim_{n\to \infty} \int_0^1 f_n(x) \,dx = 1 \neq 0 = \int_0^1 \lim_{n\to \infty} f_n(x) \,dx . \]
Double limits
(Iterated limits of a double sequence need not agree)
For \(a_{mn}\coloneqq\frac{n}{n+m}\), \[ \lim_{m\to \infty}\lim_{n\to\infty} \frac{n}{n + m} = 1 \neq 0 = \lim_{n\to \infty}\lim_{m\to\infty} \frac{n}{n + m}. \]
Suprema and limits
(The limit of the suprema need not be the supremum of the limit)
For \(f_n\coloneqq\chi_{[n,n+1]}\) on \(\RR\), \(\lim_n\sup_x\abs{f_n(x)} = 1\) and \(\sup_x\abs{\lim_nf_n(x)} = 0\).
(A pointwise limit of bounded functions need not be bounded)
On \((0,1)\), \(f_n(x)\coloneqq\min(n,1/x)\) is bounded by \(n\) and converges pointwise to \(1/x\), which is unbounded.
Continuity
(A pointwise limit of continuous functions need not be continuous)
For \(f_n(x) \coloneqq x^n\) on \([0,1]\) and \(x_k\in[0,1)\) with \(x_k\to 1\), \[ \begin{aligned} \lim_{k\to\infty}\lim_{n\to\infty} x_k^n &= \lim_{k\to \infty } 0 = 0, \\ \lim_{n\to\infty } \lim_{k\to\infty } x_k^n &= \lim_{n\to\infty} 1^n = 1, \end{aligned} \] so \(\lim_nf_n = \chi_{\theset1}\) is not continuous at \(1\).
Differentiability
(A uniform limit of differentiable functions need not be differentiable)
\(f_n(x) \coloneqq \sqrt{x^2 + {1\over n}}\) is differentiable on \(\RR\) and converges uniformly to \(\abs{x}\), which is not differentiable at \(0\).
(The limit of the derivatives need not be the derivative of the limit)
Let \(f_n(x) \coloneqq \frac{x}{1 + nx^2}\) on \(\RR\). The maximum of \(\abs{f_n}\) is \(f_n(1/\sqrt n) = \frac{1}{2\sqrt n}\), so \(f_n\to 0\) uniformly and the limit has derivative \(0\). But \[ f_n'(x) = \frac{1-nx^2}{\qty{1 + nx^2}^2}, \] so \(f_n'(0) = 1\) for all \(n\).
Likewise, for \(0<c<1\), \(g_n(x)\coloneqq n^{-c}\sin(nx)\to0\) uniformly, while \(g_n'(x) = n^{1-c}\cos(nx)\) is unbounded in \(n\) at \(x=0\).
If \(f_n\) are differentiable on an interval, \(f_n'\) converges uniformly, and \(f_n(x_0)\) converges for one \(x_0\), then \(f_n\) converges uniformly on bounded subintervals to a differentiable \(f\) with \(f' = \lim_n f_n'\). Without uniform convergence of the derivatives, a uniform limit of polynomials can be any continuous function on \([a,b]\) by the Weierstrass approximation theorem, for example a nowhere differentiable one such as the Weierstrass function.
(A differentiable function need not have a continuous derivative)
\(f(x)\coloneqq x^2\sin(1/x)\) for \(x\neq0\) and \(f(0)\coloneqq0\) is differentiable on \(\RR\), and \(f'(x) = 2x\sin(1/x)-\cos(1/x)\) has no limit at \(0\). A continuous function need not be differentiable: \(\abs x\) is continuous and not differentiable at \(0\), and the Weierstrass function is continuous and differentiable nowhere.
Integrals
(The limit of the integrals need not be the integral of the limit)
On \([0,1]\), \(f_n\coloneqq n\chi_{(0,1/n)}\to0\) pointwise while \(\int_0^1f_n = 1\).
(A pointwise limit of Riemann integrable functions need not be Riemann integrable)
Let \((q_n)\) enumerate \(\QQ\cap[0,1]\) and \(f_N\coloneqq\sum_{n\leq N}\chi_{\theset{q_n}}\). Each \(f_N\) has finitely many discontinuities and \(\int_0^1 f_N = 0\), but \(f_N\to\chi_{\QQ\cap[0,1]}\) pointwise, which is not Riemann integrable.