Let the functions $f_n$, $f$ be defined on $[0,1]$ by $f_n(x) = \begin{cases} 1 & \text{ if } x \in \left[0,{1\over 2}\right] \\ 1 - n\left(x - {1 \over 2}\right)& \text{ if } x\in \left({1\over 2}, {1\over 2}+\frac{1}{n}\right] \\ 0 & \text{ if } x\in \left({1\over 2}+\frac{1}{n}, 1\right] \end{cases} \qquad f(x) = \begin{cases} 1 & \text{ if } x\in \left[0,\frac{1}{2}\right] \\ 0& \text{ if } x\in \left(\frac{1}{2}, 1\right] \end{cases} $
Prove that $\|f_n - f\|_{\infty} = 1$ for each $n$ so that $f_n$ does not converge to $f$ in the $\sup$-norm
My Work
I have already proven pointwise convergence of $f_n \to f$, but I have shown that the $\delta$ I require for convergence to hold depends on $n$, thus implying that $f_n \not\!\to f$ uniformly. I have a theorem that says that $f_n \to f$ in the $\sup$-norm $\Longleftrightarrow$ $f_n \to f$ uniformly. However, I am wondering how to show that $\|f_n - f\|_{\infty} = 1$ for each $n$. It seems like this would take place at $\lim_{x \to 1/2^+}[f_n(x) - f(x)]$ but I am not sure how to formally show this.