3
$\begingroup$

Suppose $f$ is uniformly continuous, show $cf$ is uniformly continuous where $c$ is a real constant.

Proof: Since $f$ is uniformly continuous, we know that there exists $\delta >0$ such that $|x-y| \Rightarrow |f(x)-f(y)| < \varepsilon$ for all $x,y\in M$.

So choose $\delta = \frac{\varepsilon}{c}$, and so

$|cf(x)-cf(y)| < |cx-cy|\le|c||x-y| \le c\cdot\delta \le c\cdot\frac{\varepsilon}{c}\le \varepsilon$, thus $cf$ is uniformly continuous.

Is this proof correct?

$\endgroup$
  • $\begingroup$ No. Why $|cf(x)-cf(y)|<|cx-cy|$? $\endgroup$ – Did Oct 30 '15 at 15:27
  • $\begingroup$ That's what I was afraid of. I know I can definately factor out the c from the first term. $\endgroup$ – Michael Cera Oct 30 '15 at 15:30
3
$\begingroup$

There are a couple of mistakes:

  1. You did not say where $\epsilon$ comes from, or in other words, you forgot to add a "for each $\epsilon>0$" in the beginning of the "we know" sentence.
  2. When proving your own statement, you again did not say where $\epsilon$ comes from. Is this the "same" $\epsilon$ as in the previous time it was mentioned? It is not, is it?
  3. You should not set $\delta$ to $\frac{\epsilon}{c}$, that is not the correct way to go (first of all, because you did not specify where $\epsilon$ came from, and second of all, because this selection does not take into account the uniform continuity of $f$, so it cannot work.
  4. biggest mistake: You claim that $|cf(x)-cf(y)|<|cx-cy|$ which is completely incorrect.

What you can do is you can say that $|xf(x)-cf(y)| = |c||f(x)-f(y)|$. Then, you use the uniform continuity of $f$, along with a better selection of $\delta$, to cause $|f(x)-f(y)|$ to be small enough.

But, I advise you to first correct the first two mistakes I mention. Right now, you skipped a couple of steps, and that made you sloppy. As a beginner, never skip steps, because without experience, you don't know if what you skipped makes sense.

$\endgroup$
  • $\begingroup$ I did choose delta, but not the absolute value of c. I realized I did leave out part of the definition, it's difficult to type of a phone. $\endgroup$ – Michael Cera Oct 30 '15 at 15:36
  • $\begingroup$ @MichaelCera You chose $\delta$, but you did not choose it correctly. Also, don't type on your phone. Go home, sit behind a desk, take a piece of paper and work there. And use your computer to ask questions. $\endgroup$ – 5xum Oct 30 '15 at 15:37
3
$\begingroup$

A function $f$ is uniformly continuous on $M$ if $\forall \varepsilon>0\,\,\exists \delta>0$ such that $|x-y|<\delta\Rightarrow |f(x)-f(y)|<\varepsilon$ for all $x,y\in M$.

So choose $\delta>0$ such that for all $x,y\in M$ you have $|x-y|<\delta\Rightarrow |f(x)-f(y)|<\frac{\varepsilon}{|c|}$. Then whenever $|x-y|<\delta$ you have: $$ |cf(x)-cf(y)|\le |c|\cdot |f(x)-f(y)|<|c|\cdot\frac{\varepsilon}{|c|}=\varepsilon. $$

From this we conclude that $cf$ is uniformly continuous.

$\endgroup$
  • $\begingroup$ I fixed my previous error. $\endgroup$ – Laars Helenius Oct 30 '15 at 15:41
0
$\begingroup$

Where you wrote $|x-y| \Rightarrow |f(x)-f(y)| < \varepsilon$, you need $|x-y| < \delta \Rightarrow |f(x)-f(y)| < \varepsilon$.

You have $$ \text{For every } \varepsilon>0, \text{ there exists } \delta>0 \text{ so small that if } |x-y|<\delta \text{ then } |f(x) - f(y)|<\varepsilon. $$

This is true of EVERY positive number $\varepsilon$, and $\dfrac \varepsilon {|c|}$ is a positive number, so it is true of $\dfrac\varepsilon {|c|}$. Thus there is some $\delta>0$ so small that whenever $|x-y|<\delta$ then $|f(x)-f(y)| < \dfrac \varepsilon {|c|}$. In that case, then $|f(x)-f(y)|<\varepsilon$.

This doesn't deal with the case in which $c=0$, but it's easy to write a proof that works in that case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.