16.3 Uniform convergence of function sequences

$$\forall t, t\in \mathbb{R}\colon g_n(t)=\frac{1}{n}$$ Let $\epsilon > 0$ be given. Then choose $N=\lceil \frac{1}{\epsilon }\rceil$ . Obviously, $$\forall n, n\ge N\colon \left|g_n(t)-0\right|\le \epsilon$$ for all $t$ . Thus, $g_n(t)$ converges uniformly to $0$ .

$$\forall t, t\in \mathbb{R}\colon g_n(t)=\frac{t}{n}$$ Obviously for any $\epsilon > 0$ we cannot find a single function $g_n(t)$ for which [link] holds with $g(t)=0$ for all $t$ . Thus $g_n$ is not uniformly convergent. However we do have: $$g_n(t)\to g(t)\text{pointwise}$$