Let \(f,f_n:\mathbb R\to\mathbb R\). Assume that whenever \(x_n\to x\), \[ f_n(x_n)\longrightarrow f(x). \] Show that \(f\) is continuous.
The functions \(f_n\) are not assumed to be continuous.
Solution
1. For every fixed \(a\in\RR\),\[ f_n(a)\longrightarrow f(a). \]
Apply the hypothesis to the constant sequence \[ x_n=a, \] which converges to \(a\).
2. Let \(y_m\to x\). There exists a strictly increasing sequence of positive integers \((n_m)\) such that\[ \abs{f_{n_m}(y_m)-f(y_m)}<\frac1m \]for every \(m\).
For each fixed \(m\), step 1 applied at the point \(y_m\) gives \[ f_n(y_m)\longrightarrow f(y_m) \qquad \text{as }n\to\infty. \] Choose \(n_1\) so that the desired inequality holds for \(m=1\). Having chosen \(n_{m-1}\), choose \[ n_m>n_{m-1} \] large enough that \[ \abs{f_{n_m}(y_m)-f(y_m)}<\frac1m. \]
3. Define a sequence \((z_n)\) by\[ z_{n_m}=y_m \]for every \(m\), and\[ z_n=x \]when \(n\) is not one of the indices \(n_m\). Then\[ z_n\longrightarrow x. \]
Let \(\varepsilon>0\). Since \(y_m\to x\), there is \(M\) such that \[ \abs{y_m-x}<\varepsilon \] for all \(m\geq M\). For \(n\geq n_M\), either \(z_n=x\), or \(n=n_m\) for some \(m\geq M\), in which case \(z_n=y_m\). Thus \[ \abs{z_n-x}<\varepsilon \] for all \(n\geq n_M\).
4. For every sequence \(y_m\to x\),\[ f(y_m)\longrightarrow f(x). \]
3. Then\[ f_n(z_n)\longrightarrow f(x). \]Passing to the subsequence \(n=n_m\) gives\[ f_{n_m}(y_m)\longrightarrow f(x). \]By step 2,\[ \abs{f_{n_m}(y_m)-f(y_m)}<\frac1m\longrightarrow0. \]Therefore \(f(y_m)\to f(x)\).
5. The function \(f\) is continuous on \(\RR\).
Step 4 shows that for every \(x\in\RR\) and every sequence \(y_m\to x\), one has \[ f(y_m)\to f(x). \] By the sequential criterion for continuity in \(\RR\), \(f\) is continuous at every \(x\).
6. Q.E.D.
Step 5 is the required conclusion.