Suppose $f$ is uniformly continuous, show $cf$ is uniformly continuous. 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?
 A: 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.
A: There are a couple of mistakes:


*

*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.

*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?

*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.

*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.
A: 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.
