Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

up vote 1 down vote accepted

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 $n$.

share|improve this answer
    
Ah, thank you Ilya –  Zvpunry 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
1  
@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
add comment

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 $x = 1/2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.