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What would be an easy example of a sequence of functions defined on a compact interval so that $f_n$ goes to $f$ pointwise but $\sup f_n$ does not go to $sup f$. I thought of the usual example we take to show that the limits in integration can't be interchanged when we only have pointwise convergence. Is this correct? Does $f(x)=x^n$ work in this context? Any comments or hints? |
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Consider the functions $f_n$ defined on $[0,1]$, where $f_n$ is the function whose graph consists of the following straight line segments: from $(0,0)$ to $(1/n,1)$, from $(1/n,1)$ to $(2/n,0)$, and from $(2/n,0)$ to $(1,0)$. Note that $(f_n)$ converges pointwise to the zero function on $[0,1]$. |
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Let $f_n(x)=0$ if $x<n$ and $1$ if $x\geq n$. Then $f_n\to 0$ pointwise, but $\sup f_n = 1$ for all $n$. You can get continuous examples easily enough. |
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Let's $f:[0,1]\to \mathbb{R}$. Set $f_n(x)=x^n$ if $x<1-\frac{1}{n}$ and $f_n(x)=0$ if $x\geq 1-\frac{1}{n}$. Note that $\lim_{n\to\infty}f_n=f\equiv 0$ Then $$ \sup_{x}f_n(x)=1-\frac{1}{n} \mbox{ and } \sup_{x}f(x)=0 $$ |
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