[Math] Suppose $f$ is uniformly continuous, show $cf$ is uniformly continuous.

Question

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?

Answer 1

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.