$g_n(x)=|nx-1|;0\le x< {2\over n}$,$g_n(x)=1;{2\over n}\le x\le 1$
We need to find whether $g_n$ is uniformly convergent or not.
Could anyone tell me whether $g_n$ pointwise converges to $g$ below?
$g (x) =1$
But I am not sure whether the convergence is uniform or not.
Intuition: The pointwise limit of $g_n$ is the constant function $g(x) = 1$. The condition $x < \frac{2}{n}$ can be written as $nx < 2$, which suggests that no matter how large $n$, we can choose a small value of $x$ to make $g_n(x)$ too far away from $1$. Let's formalize this idea.
Fix $0< \epsilon < 1$. For every $N \in \mathbb N$, we want to find $n > N$ and $x \in [0, 1]$ such that $\left|g(x) - g_n(x)\right| > \epsilon$. To do this, pick any $n > N$ and pick $x \in \left(0, \frac{2}{n}\right)$ such that $\frac{\epsilon}{n} < x < \frac{1}{n}$. This is always possible since $0 < \frac{\epsilon}{n} < \frac{1}{n}$. We have: $$ \left|g(x) - g_n(x)\right| = \left|1 - |1 - nx|\right| = \left|1 - \left(1 - nx\right)\right| = nx > \epsilon $$
The second equality comes from $x < \frac{1}{n}$ and the last inequality from $\frac{\epsilon}{n} < x$.
Thus, $g_n$ doesn't converge uniformly to $g$.