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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

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

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.


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