# Proving lack of convergence in $\sup$-norm

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.

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How to show that $\|f_n-f\|_\infty=1$? Well, compute $\sup\{|f_n(x)-f(x)|\,;\,x\in[0,1]\}$. –  Did Aug 27 '12 at 12:07

## 2 Answers

Well, to show that $\|f_n - f\|_\infty = 1$ we can first mention that $|f_n(x) - f(x)|\leq 1$ for all $x\in [0,1]$ hence $\|f_n - f\|_\infty \leq 1$. To show that the equality holds, it is sufficient for any $n$ to provide a sequence $x_k$ such that $\lim\limits_{k\to\infty}|f_n(x_k) - f(x_k)| = 1$.

If you take $x_k = \frac12+\frac1k$ then $f(x_k) = 0$ and $f_n(x_k) = 1-\frac nk$ so that $$|f_n(x_k) - f(x_k)| = 1-\frac nk \to 1 \text{ with }k\to\infty$$ for any fixed $n$.

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Ah, thank you Ilya –  KingOliver Aug 27 '12 at 12:14
@jmi4: you're welcome. Note that if you have not to prove that $\|f_-f\| = 1$ and you are only looking for the lack of convergence, it would be sufficient to fix some $c>0$ and construct a sequence $y_n$ such that $|f_n(y_n) - f(y_n)|\geq c$. That may be easier to do in general - though, you have to have a guess which $c$ to take. –  Ilya Aug 27 '12 at 12:17
@Ilya, there's a typo in the last expression: the limit is 1. –  nokiddn Aug 27 '12 at 12:55
@nokiddn: thank you, fixed. –  Ilya Aug 27 '12 at 13:08

Suggestion: In problems like these, it is often very helpful to draw a picture. If you draw a picture of $f(x)$ and $f_n(x)$ for some $n$, you can see that if at any place the supremum of $|f_n(x) - f(x)|$ is achieved, it is between $1/2$ and $1/2 + 1/n$. Then concentrating on the triangle formed by the vertices $(1/2,0), (1/2,1)$ and $(1/n + 1/2,0)$ you will see that given any $n$, the triangle is not completely squashed and that the supremum is achieved exactly at $x = 1/2$.

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