Let $p: [0,1] \rightarrow R$ be a continuous function. Which of the following must be true? Justify.

1) There exists a constant $a > 0$ such that $|p(x)-p(y)|<a $ for all $x,y\in [0,1]$.

2) There exists a constant $b>0$ such that $|p(x)-p(y)|<1$ for all $x,y \in [0,1]$ that satisfy $|x-y|<b$.

3) There exists a constant $c>0$ such that $|p(x)-p(y)|<c|x-y|$ for all $x,y \in [0,1]$.

My thoughts:

1) TRUE: By the Extreme Value Theorem because [0,1] is compact, then p is bounded but i am unsure how to use this to prove the above statement.

2)TRUE: not sure why

3)FALSE: the function $p(x)= \frac{1}{x}$ is continuous on $[0,1]$ but it is unbounded.But i am unsure where to go from here.

  • 4
    $\begingroup$ "the function p(x)=1/x is continuous on [0,1]" Ouch! That one hurts... $\endgroup$ – Did Feb 16 '15 at 22:31
  1. Yep. Since $p$ is bounded, say by $M$, $|p(x)-p(y)| \leq |p(x)|+|p(y)| \leq 2M$.
  2. This is sort of like uniform continuity (which would replace $1$ by an arbitrary $\varepsilon$ and replace $b$ by $\delta$ which would depend on $\varepsilon$). Accordingly, it follows from uniform continuity, which you have because of the Heine-Cantor theorem. (A continuous function on a compact set is uniformly continuous.) It is a good exercise to prove the Heine-Cantor theorem; the easiest proof I know of uses the open cover definition of compactness.
  3. This says that the function is Lipschitz continuous, which intuitively means that it locally looks at worst piecewise linear. So if you have a function with a vertical tangent line somewhere, it will fail. A family of examples is given by $f(x)=x^a$ on $[0,1]$ when $0<a<1$.

Note that in your attempt at the third part, $f(x)=1/x$ is not continuous on $[0,1]$, in fact it is not even defined on $[0,1]$.

The following is a bit pedantic and not needed to solve the problem. I bring it up because I don't like making a statement which is false, even if it is to provide intuition.

The Lipschitz property is slightly weaker than the intuition I suggested. For instance, we could have a nasty function on $[0,1]$ which is:

  • $0$ at zero,
  • linear with slope $1$ on each interval $(1/(n+1),1/n)$ when $n$ is odd,
  • linear with slope $-1$ on each interval $(1/(n+1),1/n)$ when $n$ is even, and
  • continuous at the remaining points.

Such a function doesn't look locally piecewise linear near $0$ (it changes direction infinitely many times even on a small interval near $0$), but it is still Lipschitz. My point about a vertical tangent line still goes through, however.


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