Cfrgtkky

Let $f: R rightarrow R$ be a differentiable function. Prove that if $f'$ is bounded, then $f$ is uniformly continuous

My attempt:

Since $f$ is differentiable, we have $f'(x) = lim_{x to x_0}frac{f(x) - f(x_0)}{x-x_0}$.

Since $f$ is differentiable, it follows that $f$ is also continuous. Therefore, given $epsilon > 0$, there is some $delta > 0$ such that $|x-x_0|<delta rightarrow |f(x)-f(x_0)|<epsilon$.

$frac{|x-x_0|}{|x-x_0|}*|f(x)-f(x_0)|=|x-x_0|frac{|f(x)-f(x_0)|}{|x-x_0|}<deltafrac{|f(x)-f(x_0)|}{|x-x_0|}<epsilon$

Now take $lim_{x to x_0}$ of both sides to get: $delta f'(x) < epsilon$. Since $f'$ is bounded, there is some $M$ such that $-M leq f' leq M$. So we have $delta f'(x)leqdelta M<epsilon$. Take $delta = frac{epsilon}{M}$. Therefore, $f$ is uniformly continuous.

Is my logic here correct? I am not sure if I can do the limit step.

The $delta$ you end up choosing is good; this $delta$ will work. The problem is your justification why it works. It almost works with a bit of unpacking, but it needs to be rewritten in a way that is clear flows logically from the assumption that $|x - x_0| < delta = varepsilon/M$ to the conclusion that $|f(x) - f(x_0)| < varepsilon$.

You write

Since $f$ is differentiable, it follows that $f$ is also continuous. Therefore, given $varepsilon > 0$, there is some $delta > 0$ such that $|x-x_0|<delta rightarrow |f(x)-f(x_0)|<varepsilon$.

This is true, but it's a confusing step to put in your proof. The way I interepeted this when I first read it is that you are choosing a value for the variable $delta$. You don't know what it is, but you know one exists that satisfies the above property (given the fixed $varepsilon > 0$ and $x_0 in mathbb{R}$). The rest of your proof would then (somehow) have to justify why this $delta$ does not depend on $x_0$ at all, or perhaps define another $delta'$, based on $delta$, that didn't depend on $x_0$. Since you end up simply defining a totally different $delta$, not relating to this $delta$, the step is confusing, and probably should be dropped.

Next, you write

$frac{|x-x_0|}{|x-x_0|} times |f(x)-f(x_0)|=|x-x_0|frac{|f(x)-f(x_0)|}{|x-x_0|}<deltafrac{|f(x)-f(x_0)|}{|x-x_0|}<varepsilon$

This is another reason why the previous step is confusing; you are now using $delta$ in inequalities. Without first defining $delta$, this is a bit confusing. You haven't talked about what you're assuming here. For example, it appears that you're assuming $|x - x_0| < delta$. If you have a $delta$ defined (such as from the previous paragraph), this is a very reasonable assumption to make, but it should be made explicit before writing this step.

It's also not clear why $deltafrac{|f(x)-f(x_0)|}{|x-x_0|}<varepsilon$. Without a clear definition of $delta$, I don't know why this would be true.

You also write

Now take $lim_{x to x_0}$ of both sides to get: $delta f'(x) < epsilon$.

This is a small point: if you have $g(x) < N$ for all $x$, then $lim_{x to a} g(x) le N$. You cannot conclude strict inequality!