Failure of Property Preservation under Pointwise Convergence

Construction. Define, for each $n \in \mathbb{N}$,

Step 1: Each $f_n$ is continuous on $[0,1]$. The function $x \mapsto x^n$ is a polynomial (a finite product of the identity function with itself $n$ times, composed with finite sums), and every polynomial is continuous on $\mathbb{R}$, hence on $[0,1]$. (This uses only the algebra-of-limits characterization of continuity: sums and products of continuous functions are continuous, and constants and $x \mapsto x$ are continuous.)

Step 2: Identify the pointwise limit. Fix $x \in [0,1]$ and compute $\lim_{n\to\infty} x^n$.

• If $x = 1$: $f_n(1) = 1^n = 1$ for every $n$, so $f_n(1) \to 1$.
• If $0 \le x < 1$: we claim $x^n \to 0$. Given $\varepsilon > 0$ with $\varepsilon < 1$ (WLOG, since larger $\varepsilon$ is even easier), if $x = 0$ this is trivial ($f_n(0)=0$ for all $n\ge 1$). If $0 < x < 1$, write $x = \frac{1}{1+h}$ with $h > 0$ (possible since $0<x<1$ forces $1/x > 1$, so $h := 1/x - 1 > 0$). By Bernoulli's inequality, $(1+h)^n \ge 1 + nh$ for all $n \ge 1$ (proved by induction: true at $n=1$; if $(1+h)^n \ge 1+nh$ then $(1+h)^{n+1} = (1+h)^n(1+h) \ge (1+nh)(1+h) = 1+(n+1)h+nh^2 \ge 1+(n+1)h$, using $h>0$). Hence

since $1/(1+nh) \to 0$ by the Archimedean property applied to choose $N$ with $N > \frac{1}{\varepsilon h}$, giving $\frac{1}{1+Nh} < \varepsilon$. So by the squeeze principle, $x^n \to 0$. Therefore the pointwise limit function is

This defines $f$ at every point of $[0,1]$, so $(f_n) \to f$ pointwise on $[0,1]$.

Step 3: $f$ is discontinuous at $x = 1$. We must show it is false that $\lim_{x \to 1^-} f(x) = f(1)$. Take the sequence $x_k = 1 - \frac{1}{k}$ for $k \ge 2$; then $x_k \to 1$ and $x_k \in [0,1)$ for all such $k$, so $f(x_k) = 0$ for every $k$, giving $f(x_k) \to 0 \ne 1 = f(1)$. Since continuity at $1$ would require every sequence $x_k \to 1$ in $[0,1]$ to satisfy $f(x_k) \to f(1)$ (sequential characterization of continuity), and we exhibited one that fails this, $f$ is discontinuous at $x=1$. $\blacksquare$

Construction. For $n \ge 1$ define the "moving triangular spike"

Step 1: Each $g_n$ is continuous on $[0,1]$. On each of the three closed subintervals $[0,\tfrac1n]$, $[\tfrac1n,\tfrac2n]$, $[\tfrac2n,1]$ (assuming $n \ge 2$ so that $\tfrac2n \le 1$; for $n=1$ the middle piece runs to $x=2$ but we simply restrict to $[0,1]\subset[0,2]$, and the argument is identical, only the third piece is vacuous), $g_n$ is given by a linear (hence continuous) formula. At the junction $x = 1/n$: the left formula gives $n^2 \cdot \frac1n = n$, and the right formula gives $n^2\left(\frac2n - \frac1n\right) = n^2 \cdot \frac1n = n$; these agree. At $x = 2/n$ (for $n\ge 2$): the middle formula gives $n^2\left(\frac2n-\frac2n\right)=0$, matching the third piece's value $0$. By the pasting lemma for continuity (a function built from finitely many continuous pieces on closed sets, agreeing at the shared endpoints, is continuous on the union), $g_n$ is continuous on all of $[0,1]$.

Step 2: The pointwise limit is identically $0$. Fix $x \in [0,1]$.

• If $x = 0$: $g_n(0) = 0$ for every $n$ (from the first formula, $n^2\cdot 0 = 0$), so $g_n(0) \to 0$.
• If $x \in (0,1]$: by the Archimedean property, choose $N$ such that $\frac{2}{N} < x$. Then for every $n \ge N$ we have $\frac{2}{n} \le \frac{2}{N} < x$, so $x$ lies in the region $\left[\frac2n, 1\right]$ where $g_n(x) = 0$ by definition. Hence $g_n(x) = 0$ for all $n \ge N$, so trivially $g_n(x) \to 0$.

In both cases $g_n(x) \to 0$, so $(g_n) \to g \equiv 0$ pointwise on $[0,1]$, and $\int_0^1 g(x)\,dx = \int_0^1 0\,dx = 0$.

Step 3: Compute $\int_0^1 g_n(x)\,dx$ exactly for every $n$. The graph of $g_n$ on $[0, 2/n]$ is a triangle with base $2/n$ and height $g_n(1/n) = n$ (computed above), and $g_n = 0$ outside $[0,2/n] \cap [0,1]$. (For $n \ge 2$ the full triangle lies in $[0,1]$; for $n=1$ we only integrate over $[0,1]$, which is the left half of the triangle — but let us just do the computation directly by the fundamental theorem of calculus / elementary area formula for $n\ge 2$, and note $n=1$ doesn't affect the limit.) For $n \ge 2$:

(This triangle-area computation can be justified rigorously by splitting the integral at $1/n$ and computing each linear piece via the Fundamental Theorem of Calculus: $\int_0^{1/n} n^2x\,dx = n^2\cdot\frac{(1/n)^2}{2} = \frac12$, and $\int_{1/n}^{2/n} n^2\left(\frac2n - x\right)dx = n^2\left[\frac{2}{n}x - \frac{x^2}{2}\right]_{1/n}^{2/n} = \frac12$ by direct antiderivative evaluation; summing gives $1$.) So for every $n \ge 2$,

Step 4: Conclude. Hence $\lim_{n\to\infty} \int_0^1 g_n(x)\,dx = \lim_{n\to\infty} 1 = 1$, while $\int_0^1 g(x)\,dx = 0$. These are unequal ($1 \ne 0$), so

even though $g_n \to g$ pointwise on all of $[0,1]$ and every $g_n$ is continuous (indeed Riemann integrable). $\blacksquare$

Construction. For $n \ge 1$ define

Step 1: Each $h_n$ is differentiable with continuous derivative. $h_n$ is a composition/scalar multiple of the sine function, which is differentiable everywhere with $\frac{d}{dx}\sin(u) = \cos(u)\cdot u'$ (chain rule, a standard consequence of the definition of the derivative applied to $\sin$). Hence

and this is continuous since $\cos$ is continuous and composition/scalar multiplication of continuous functions is continuous.

Step 2: The pointwise limit is $h \equiv 0$. Fix $x \in \mathbb{R}$. Since $|\sin(n^2 x)| \le 1$ for all $n$ and all $x$ (boundedness of sine, a basic property from its definition via the unit circle or its power series), we get

as $n \to \infty$, by the Archimedean property (for any $\varepsilon>0$ choose $N > 1/\varepsilon$). By the squeeze principle, $h_n(x) \to 0$ for every fixed $x$. So $(h_n) \to h$ pointwise on $\mathbb{R}$, where $h(x) = 0$ for all $x$.

Step 3: $h$ is differentiable everywhere, with $h'(x) = 0$ for all $x$. This is immediate: the derivative of the identically-zero function is identically zero, directly from the definition

Step 4: Show $h_n'(x) \not\to h'(x)$ at a specific point. Take $x = 0$. Then

So $h_n'(0) = n \to \infty$ as $n \to \infty$ — in particular $(h_n'(0))$ does not converge to $h'(0) = 0$; it does not converge at all (it is unbounded, hence has no finite limit — if it converged to some $L$ it would be bounded, contradicting $h_n'(0) = n \to \infty$). Thus at $x=0$,

(indeed the left side does not even exist as a finite number), despite $h_n \to h$ pointwise everywhere and every $h_n$ being smooth with continuous derivative. $\blacksquare$