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If $f(x)$ takes a finite real value for all $x$ on the closed interval $[a,b]$, must there be a real number $M$ such that $M\geq f(x)$ for all $x$ on this interval? It seems that if not, there must be a point $c\in[a,b]$ such that $\lim_{x\to c}f(x)=+\infty$, and so $f(x)$ must be undefined at some point on this interval, but I don't know how to make this rigorous.

Edit: I see that $f(0)=0$, $f(x)=1/x$ on $(0,1]$ is a counterexample. I also see that I have been imprecise with terminology. Let me modify the question: Is there always a sub-interval $[a',b']$ with $a<a'<b'<b$ such that there is an upper bound for $f(x)$ on this sub-interval?

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  • $\begingroup$ Your title is different from the question in the post: $M\ge f(x)$ does not mean that $M$ is maximum, it is an upper bound. In the other words, if such $M$ does not exist, $f$ is not bounded from above. $\endgroup$ – Martin Sleziak Nov 2 '12 at 15:44
  • $\begingroup$ If $f$ is continuous, then it must be bounded, see ProofWiki. It seems that in your post you use the assumption $f(c)=\lim\limits_{x\to c}f(x)$, which is an equivalent condition for continuity of a function $f \colon \mathbb R \to \mathbb R$. $\endgroup$ – Martin Sleziak Nov 2 '12 at 15:46
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How about: $f(x)$ on $[0, 1]$ where $f(0) = 0$ and, for all other points, $f(x) = 1/x$.

Edit: Since you added an extra part to your question, you could use the following function:

$f(x) = x$ if $x$ is irrational;

$f(x) = n$ if $x$ is rational in lowest terms, where $n:=$ highest power of $2$ dividing $x$'s denominator.

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About your modified questions:

Is there always a sub-interval $[a',b']$ with $a<a'<b'<b$ such that there is an upper bound for $f(x)$ on this sub-interval?

There are real functions, which are unbounded on each interval.

See Function with range equal to whole reals on every open set at MO. Many functions from this question work too: Can we construct a function $f:\mathbb R\to \mathbb R$ such that it has intermediate value property and discontinuous everywhere?. Discontinuous solutions of Cauchy functional equation have this property, see e.g. here.

To give a very simple example, we define $f \colon [0,1] \to \mathbb R$ by $$f(x)= \begin{cases} 0; & x\notin\mathbb{Q},\\ q; & x=\frac{p}{q}; \gcd(p,q)=1; p\ne 0,\\ 1; & x=0. \end{cases}$$ (We took reciprocals of non-zero values of Thomae's function.)

For each prime number $q$ if you take interval of length greater then $\frac2q$ then this interval will contain an $x$ such that $f(x)=q$.

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  • $\begingroup$ Thank you - I never would've thought of such crazy functions. $\endgroup$ – jcai Nov 2 '12 at 16:05

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