Winter 2004 #4

Problem Statement: Suppose that f(x) is differentiable on the bounded interval (a,b). Suppose further that there is a constant M such that |f'(x)|<M for every x\in (a,b). Show that lim_{x\rightarrow a}f(x) exists and is finite.

Note: When I read this question the first time I thought the qual writers were crazy because I found a counter example. If we have a piecewise smooth function on \mathbb{R} that is continuous on (a,b) with a jump discontinuity for x=a then the lim_{x\rightarrow a}f(x) doesn’t have to exist since the right and left handed limits are not equal. Done! But, after much deliberation I’ve concluded that the qual writers are not crazy and they must really be asking me to show that lim_{x\rightarrow a^+}f(x) exists and is finite. So, with that in mind I was able to continue.

Proof:

Since f(x) is differentiable on (a,b) it follows that f(x) is continuous on (a,b). Let M>0 such that \left|f'(x)\right|<M for every x\in (a,b).

Claim: f(x) is uniformly continuous on (a,b). Let x,y\in (a,b) and wlog x<y. Then f(x) is continuous and differentiable on [x,y], so we may apply the Mean Value Theorem, which states that there exists a c\in(x,y) such that f'(x)=\dfrac{f(y)-f(x)}{y-x}. By our assumption it follows that \left| f'(c)\right|=\left|\dfrac{f(y)-f(x)}{y-x}\right|=\dfrac{\left|f(y)-f(x)\right|}{\left|y-x\right|}<M. This implies that \left|f(y)-f(x)\right|<M\left|y-x\right|. Let \varepsilon >0 and \delta=\dfrac{\varepsilon}{M}. Then for |y-x|<\delta it follows that |f(y)-f(x)|<M|y-x|<M\left(\dfrac{\varepsilon}{M}\right)=\varepsilon. Since x and y were arbitrary in (a,b) it follows that f is uniformly continuous on (a,b).

Let \{x_n\} be a sequence converging to a such that x_n\in (a,b) for every n. Since this is a convergent sequence we get that \{x_n\} must be Cauchy. Let \varepsilon'=\delta where \delta is as defined above. Then since \{x_n\} is Cauchy there exists and N\in \mathbb{N} such that |x_n-x_m|<\varepsilon'=\delta whenever n,m>N. Since \varepsilon'=\delta this implies that |f(x_n)-f(x_m)|<\varepsilon. Thus, \{f(x_n)\} is Cauchy and so \{f(x_n)\} converges as \{x_n\}\rightarrow a. Therefore \lim_{x_n\rightarrow a}f(x_n) exists and is finite. But this is equivilent to the statement that the \lim_{x\rightarrow a}f(x) exists and is finite.

\Box

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This entry was posted in Analysis, Cauchy, Continuity, Differentiable, Math, MVT, Sequence, Uniform Continuity and tagged , , . Bookmark the permalink.

4 Responses to Winter 2004 #4

  1. Richard says:

    This is a good problem, it uses two of my favorite things from analysis: mean value theorem and Cauchy sequences! Bounded derivative implies uniform continuity, that’s a classic application of mvt.

  2. DR says:

    I think it’s not done yet! You have shown that the image sequence of any Cauchy sequence in the domain that tends to a converges to something. You must argue that they _all_ tend to the same limit L. That common value is then the limit of f(x) as x tends to a.

    Keep up the good work…

    • eekelly2388 says:

      So, I think right before the second to last sentence I should add “Let L=\lim_{n\rightarrow\infty}f(x_n). Suppose \{y_n\} is another sequence in (a,b) convering to a. Then \{f(x_n)\} and \{f(y_n)\} must converge to the same limit, L, as n\rightarrow \infty since limits are unique. ”

      Is that enough to say or should I go into that argument in more detail?

      Thanks for the feedback!

  3. Suzanne Lavertu says:

    I want to note that we need f(x) to be uniformly continuous on (a,b) so that we can obtain the inequality, |f(x_n)-f(x_m)| < \epsilon when |x_n-x_m| < \delta, so that our sequence (f(x_n)) is Cauchy and therefore convergent.

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