Consider the following probability density function $f(x)$
\begin{cases} 0 & x<-1 \\ 1+x & z\in[-1,0] \\ 1-x & z\in[0,1] \\ 0 & x>1 \end{cases} Then the characteristic function is calculated as $$\varphi(t)=\mathbb{E(e^{itx})}=\int_{\mathbb{R}}e^{itx}f(x)dx=\int_{-1}^0e^{itx}(1+x)dx+\int_0^1e^{itx}(1-x)dx$$ Which, according to Wolfram Alpha, yields $$\varphi(t)=\frac{2-e^{it}-e^{-it}}{t^2}$$ How is this possible? The characteristic function is supposed to be uniformly continuous in $\mathbb{R}$ while this one obviously fails for $t=0$.