question about uniform continuity

Question

I have a function $f:[a, b]\rightarrow \mathbb{R}^n$ such that $f$ is piecewise continuous in the intervals $[a, t_1],$ $(t_1, t_2],$ $(t_2, t_3],$...., $(t_n, b].$ Can we say that $f$ is uniformly piecewise continuous on $[a, b]?$

Answer 1

No. A counter example is $f: [-1,1] \to \mathbb R$ defined as

$$f(x)=\begin{cases}0, &-1 \leq x \leq 0 \\ \frac 1x, &0<x\leq 1 \end{cases}$$

Answer 2

Even if the function is left/right continuous at each $\;t_k\;$ but not continuous necessarily at all of them you will lose unif. continuity:

$$f(x)=\begin{cases}x,&0\le x<1\\{}\\7,&1\le x\le2\end{cases}$$

Thus, for example, you can take

$$\begin{cases}\{x_n\}=\left\{\frac{n-1}n\right\}\subset[0,2]\\{}\\\{y_n\}=\{1\}\subset[0,2]\end{cases}\;\;\;\implies |y_n-x_n|=\frac1n\xrightarrow[n\to\infty]{}0\;,\;\;\text{yet}$$$${}$$

$$|f(y_n)-f(x_n)|=\left|7-\frac{n-1}n\right|=\frac{6n+1}n\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0$$

and we're done.