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Let $f$ be a nonzero continuous function on $\mathbb R$. If we write $\sup_{x\in \mathbb R}|f(x)|=c$, does it mean that there is (at least) one $x_{o}\in \mathbb R$ such that $|f(x_{o})|=c$? Just to be sure!

Edit: $|f(x_{o})|=c$ instead of $f(x_{o})=c$, my false:)

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Consider $f:\mathbb{R}\to \mathbb{R}$ defined by


$\sup{|f(x)|}=\frac{\pi}{2}$, but $f$ is strictly increasing function and there is no $x$ such that $|f(x)|=\frac{\pi}{2}$.

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@Gasto'n: If $\lim_{x\to\pm\infty}|f(x)|=0$ then we could find such point $x_{o}$, is that correct? (I already asked this question before but with no clear answer!!). – Catherine Jun 17 '12 at 1:30
@Catherine Yes, because in that case, $f(x) < \frac{c}{2}$ outside an interval $[-a,a]$, and $f(x)$ attains it's sup on $[-a,a]$. – N. S. Jun 17 '12 at 1:32
@Catherine: You need to be careful: it's possible that $f$ never takes any value anywhere near $c$, because you are taking absolute values when considering the supremum, but not when considering the function. E.g., take $f(x) = -e^{-x^2/2}$. Then $\sup_{x\in\mathbb{R}}|f| = 1$, but $f$ is always negative. In the situation when the limits equal $0$, you can find $x$ where $|f(x)|=c$, but you don't know whether $f(x)=c$ or $f(x)=-c$. – Arturo Magidin Jun 17 '12 at 1:35
@N.S.: Is this an answer for my comment to Gasto'n, or to my original post above? – Catherine Jun 17 '12 at 1:36
@Catherine I didn't understand your second question, are you adding an aditional condition to original problem? In this case the answer is yes since $\mathbb{R}$ is complete and $|f([-a,a])|$ must be bounded and take maximum value and $|f|$ is also continuous (where $a$ is such that $|f(x)|<0.9c$ in $[-a,a]$). – Gastón Burrull Jun 17 '12 at 1:44

There are two reasons why this can fail:

  1. $|f|$ does not achieve its maximum; this occurs in the example given by Gastón Burrull.

  2. The supremum of $|f|$ may not be the supremum of $f$. For example, even though $f(x)=-\exp(-x^2/2)$ is such that $|f|$ achieves its maximum (which is $1$ at $x=0$), there is no point where $f(x)=1$ because $f(x)\lt 0$ for all $x$.

So, first, your conclusion should be "there exists $x$ such that $|f(x)|=c$." The conclusion will follow in several situations; for example, it will follow if $$\lim_{x\to\infty}|f(x)|\lt c \qquad \text{and}\qquad \lim_{x\to-\infty}|f(x)|\lt c.$$ Indeed, assume that this is the case: let $R=\lim\limits_{x\to\infty}|f(x)|$ and $L=\lim\limits_{x\to-\infty}|f(x)|$. Then there exists $N\gt 0$ such that for all $x\gt N$ $\Bigl| |f(x)|-R\Bigr|\lt \frac{1}{2}(c-R)$, and there exists $M\gt 0$ such that for all $x\lt -M$, $\Bigl| |f(x)|-L\Bigr|\lt \frac{1}{2}(c-L)$. Then for all $x\notin [-M,N]$, we have $|f(x)|\lt c - \frac{1}{2}\min\{c-L, c-R\}$, so $|f(x)|$ does not get arbitrarily close to $c$ on $(-\infty,M)\cup (N,\infty)$. Thus, the supremum must be achieved inside the finite closed interval $[-M,N]$, and since $|f(x)|$ is continuous, the supremum is actually achieved, so there exists $a\in [-M,N]$ such that $|f(a)|=c$.

(The case where $\lim\limits_{x\to\infty}f(x) = \lim\limits_{x\to-\infty}f(x) = 0$ is a special case of the above when $f$ is not the constant function $0$).

In fact, a weaker condition is that the result will follow if there exists $N\gt 0$ such that $\sup_{|x|\gt N}|f(x)|\lt c$, which is really all we need above to reduce to a bounded interval where $f$ must take the value $c$. And also under the even weaker condition that there exist $N\gt 0$ such that $\sup\{|x|\leq N\} |f(x)| = c$.

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A minus sign in definition of $L$. – Gastón Burrull Jun 17 '12 at 1:57
@GastónBurrull: Thanks! – Arturo Magidin Jun 17 '12 at 1:58

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