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Let $~f: \mathbb{R} \to \mathbb{R}$ be bounded and uniformly continuous. Show that $e^f$ is also bounded and uniformly continuous.

My approach (with 5xum's extensive help) goes as follows:

  1. $f$ bounded $\implies$ $e^f$ bounded

Since there exists an $M \in \mathbb{R}_{>0}$ so that $|f(x)| \leq M$ for all $x \in \mathbb{R}$, by the continuity and monotonicity of $e^x: \mathbb{R} \to \mathbb{R}$, we can see that $|e^{f(x)}| = e^{f(x)} \leq e^M ~\forall x \in \mathbb{R}$.


  1. $f$ uniformly continouos. $\implies$ $e^f$ uniformly continuous.

We want to show that $$ \forall \varepsilon' > 0 ~\exists \delta' > 0 ~\forall x,y \in \mathbb{R}: (|x-y| < \delta' \implies |e^{f(x)} - e^{f(y)}| < \varepsilon') $$ Let $\varepsilon' > 0$ be fixed and set $\varepsilon := \frac{\varepsilon'}{2 M} > 0$, where $M$ is the bound of $\max\{e^{f(x)},e^{f(y)}\}$, meaning $|e^{f(x)}|, |e^{f(y)}| \leq M ~\forall x,y \in \mathbb{R}~~(*_2)$.

Since $f$ is uniformly continuous, the following holds $$ \forall \varepsilon > 0 ~\exists \delta > 0 ~\forall x,y \in \mathbb{R}: |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon ~~(*_1) $$

Now, let $\delta > 0$ be so that $|x - y| < \delta$ and without loss of generality let $f(y) \geq f(x) ~\forall x,y\in\mathbb{R}$. Then we obtain $$ \begin{align*} |e^{f(x)} - e^{f(y)}| = |e^{f(x)}| |1 - e^{f(y) - f(x)}| &= e^{f(x)} |e^{|f(y) - f(x)|} - 1| \\ &\overset{(*)}{<} 2 e^{f(x)} |f(x) - f(y)| \\ &\overset{(*_1)}{<} 2 e^{f(x)} \varepsilon \\ &\overset{(*_2)}{\leq} 2 \varepsilon M \\ &=\varepsilon' \end{align*} $$

Where $(*)$ can be done because $|e^u - 1| < 2u ~~\forall u \in (0,1]$.

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  1. $f$ bounded $\implies$ $e^f$ bounded

trivial because of the monotonicity of $e^x$ ??

My professor had a favourite saying: "Dear colleague, 'trivial' and 'obvious' are my words. They are not yours to use."

If what you wrote is so trivial, then you should be able to prove it in $3$ lines, and if you are able to prove in $3$ lines, then just do it.

Beside, $f(x)=\frac1{x-1}$ is a monotonic function on $(0,1)$, but it is not bounded, so your argument obviously needs work.


  1. $f$ uniformly cts. $\implies$ $e^f$ uniformly cts.

We want to show that $$ \forall \varepsilon' > 0 ~\exists \delta' > 0 > \forall x,y \in \mathbb{R}: (|x-y| < \delta' \implies |f(x) - f(y)| < > \varepsilon') $$

No. We assume that what you wrote is true. What we need to prove is the following:

$$\forall\epsilon >0 \exists \delta > 0 \forall x,y\in\mathbb R:(|x-y|<\delta\implies |e^{f(x)} - e^{f(y)}| < \epsilon)$$

You need to prove the uniform continuity of $e^f$, while assuming that $f$ is uniformly continuous and bounded.


After your edit:

Part 1 is ok now.

Part 2 is still strange.

First problem:

Let $\varepsilon > 0$ be fixed and $\varepsilon' > 2 M \varepsilon >0$

This is very strange. Your proof should start with "Let $\varepsilon' > 0$", not "$\varepsilon > 0$." I think this is only a consequence of sloppy writing, though, so you should write, instead of the above,

Let $\varepsilon' > 0$ be fixed and set $\varepsilon = \frac{\varepsilon'}{2M}$

And then the proof should work out.

Second problem:

From $|x - y| < \delta$ follows that $|f(x) - f(y)| < \varepsilon$ because $f$ is uniformly cts.

This is very sloppy writing. What is $\delta$ equal to here? You haven't defined $\delta$, it just appeared out of the blue. This is another example of sloppy writing that must be fixed.

Third problem:

$$|e^{f(x)} - e^{f(y)}| = |e^{f(x)}| |e^{|f(x) - f(y)|} - 1|$$

This is false if $f(x)<f(y)$. For example, if $f(x)=0$ and $f(y)=1$, then the left side of the equation equals $|e^0-e^1|=e$, while the right side equals $|e^0||e^{|0-1|} - 1| = 1\cdot |e-1|=e-1$

Fourth problem:

$$|e^{f(x)}| |e^{|f(x) - f(y)|} - 1| < e^{f(x)} |e^{\varepsilon} - 1|$$

How do you know this inequality is true? If I were you, I would first use the knowledge you write in $*$ before going for the estimate with $\delta$...


After your second edit:

Slight problem:

Don't say "let $\delta$ be fixed", better to just say "let $\delta$ be such that if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$. That way, it's clear you only fix one number ($\epsilon$) and then "calculate" what $\delta$ is.

Bigger mistake:

You still wrote

$$|e^{f(x)}| |1 - e^{f(y) - f(x)}| = e^{f(x)} |e^{|f(y) - f(x)|} - 1|$$

which is not true, since, for $f(x)=1$ and $f(y)=0$, the left side equals $$|e^1|\cdot |1-e^{0-1}| = e\cdot (1-e^-1) = e-1$$

while the right side equals

$$|e^1|\cdot |e^{|0-1| -1} = e\cdot (e-1)=e^2-1$$

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  • $\begingroup$ I wasn't sure how to prove the boundness (hence the ? at the end), the 'trivial'-statement was a hint received from a friend. I now added a proof to my question, is it correct? And you're right about the "We want to show"-part. I misspelled and now edited my question. $\endgroup$ – Viktor Glombik Apr 10 '18 at 12:03
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    $\begingroup$ @ViktorGlombik Your proof is very very sloppy. You obviously have the correct idea, but if I was grading this, you would struggle to get half the points. There are at least 4 problems with your proof which I described in my edited answer. $\endgroup$ – 5xum Apr 10 '18 at 12:22
  • $\begingroup$ I've tried to incorporate your suggestions. Is my proof satisfactory now? $\endgroup$ – Viktor Glombik Apr 10 '18 at 13:00
  • $\begingroup$ @ViktorGlombik Unfortunatelly, no. It's much closer, but still sloppy. $\endgroup$ – 5xum Apr 10 '18 at 13:09
  • $\begingroup$ but I said that log $f(y) \geq f(x)$, so that $f(y) - f(x) = |f(y) - f(x)|$. In your counterexample, you ignore that condition, or am I mistaken? $\endgroup$ – Viktor Glombik Apr 10 '18 at 13:29

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