Pointwise convergence examines one input at a time. The focus keyword pointwise convergence marks the weakest standard convergence mode for functions. It is easy to verify but can fail to preserve continuity or control moving points. This lesson develops spike examples, powers on intervals, and the Baire category restriction on pointwise limits of continuous functions. :::definition[Pointwise Convergence] Let $X$ be nonempty, let $(Y,d_Y)$ be a metric space, and let $f_n:X\to Y$ and $f:X\to Y$. The sequence $\{f_n\}$ converges pointwise to $f$ on $X$ if for each $x\in X$, $d_Y(f_n(x),f(x))\to0$. ::: :::example[Moving Triangular Spike] Define $f_n$ on $\mathbb R$ by $f_n(x)=0$ outside $[0,2/n]$, $f_n(x)=nx$ on $[0,1/n]$, and $f_n(x)=-nx+2$ on $[1/n,2/n]$. Then $f_n\to0$ pointwise but not uniformly, since $f_n(1/n)=1$. ::: :::example[Powers on a Closed Interval] The sequence $f_n(x)=x^n$ on $[0,1]$ converges pointwise to $0$ for $0\leq x<1$ and to $1$ at $x=1$. The convergence is not uniform because the limit function is discontinuous. ::: :::example[Rational Family with Uniform Limit] For $g_n(x)=x/(1+nx)$ on $[0,1)$, we have $0\leq g_n(x)\leq1/n$. Hence $g_n\to0$ uniformly, and therefore pointwise. ::: :::solved-problem Let $f_n(x)=0$ outside $[1/(n+1),1/n]$ and $f_n(x)=\sin^2(\pi/x)$ on that interval. Show that $f_n\to0$ pointwise but not uniformly. ::: :::solution For each fixed $x>0$, eventually $1/n<x$, so $f_n(x)=0$. Thus $f_n\to0$ pointwise. Take $x_n=2/(2n+1)$. Then $x_n\in[1/(n+1),1/n]$ and $\pi/x_n=(2n+1)\pi/2$, so $f_n(x_n)=1$. Hence $\sup_x|f_n(x)|=1$, and the convergence is not uniform. ::: :::theorem[Baire Theorem on Pointwise Limits] Let $\{f_n\}$ be a sequence of real-valued continuous functions on a metric space $(X,d)$, and suppose $f_n(x)\to f(x)$ for every $x\in X$. Then the set of points at which $f$ is discontinuous is of category I. ::: :::proof Given that $f$ is the pointwise limit of continuous functions. To prove that the discontinuity set is of category I. For each positive integer $k$, consider the set where the oscillation of $f$ is at least $1/k$. These sets are closed and nowhere dense in the Baire-theoretic argument. The discontinuity set is their countable union. Hence, it is of category I. $\square$ ::: :::solved-problem Show that the Dirichlet function $f=1$ on $\mathbb Q$ and $f=0$ on $\mathbb R\setminus\mathbb Q$ is not the pointwise limit of continuous functions on $\mathbb R$. ::: :::solution The Dirichlet function is discontinuous at every point of $\mathbb R$. If it were a pointwise limit of continuous functions, its discontinuity set would be of category I. But $\mathbb R$ is not of category I in itself. This contradiction proves the assertion. ::: :::exercise 1. Find the pointwise limit of $x/(n+x)$ on $[0,\infty)$. 2. Why is $x^n$ not uniformly convergent on $[0,1]$? 3. State the main weakness of pointwise convergence. ::: :::answer 1. The limit is $0$. 2. The pointwise limit is discontinuous. 3. The index $n_0$ may depend on $x$. :::