I was reading the proof of a theorem and at a point of the argument the author just posed an inequality which I was unable to derive, but is probably just a smart application of the triangle inequality.

The problem is as follows. Let $f,g:\mathbb{R} \to \mathbb{R}$ be uniformly continuous and $\epsilon, \delta >0$ such that if $x,y\in \mathbb{R}$ are st $|x-y| < \delta$, that then $|f(x)-f(y)|, |g(x)-g(y)| <\epsilon$. Let $I \subset \mathbb{R}$ be an interval with $diam(I) < \delta$

The author then states without proof that $\sup_{x,y\in I} |f(x)-g(y)|^p < |f(t)-g(t)+2\epsilon|^p$ for all $t \in I$., $p\geq 1$

It feels as if this should be easy to prove, but I have been stuck at it for quite some time now. Any help would be greatly appreciated.

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    $\begingroup$ I edited the problem formulation to the one as in the proof as there was a crucial mistake in my original formulation. Sorry! $\endgroup$
    – pokemonfan
    Mar 6, 2019 at 22:04

2 Answers 2


It appears to be a mistake, possibly a typo. Consider the following $2$ uniformly continuous functions

$$f(x) = bx - c \tag{1}\label{eq1}$$ $$g(x) = \left| x \right| + a \tag{2}\label{eq2}$$

with $a, b \in \mathbb{R}$, $a \gt 0$, $b \gt 1$ and $c \gt 0$. Also, let $\delta = 1$. In this case, $|f(x)-f(y)| \lt b$ and $|g(x)-g(y)| \lt 1$ in the interval, so we can choose $\epsilon = b$. Next, let $I = [-0.9, -0.1]$. Note that

$$\left| f(-0.9) - g(-0.1) \right| = \left| -0.9b - c - 0.1 - a \right| = 0.9b + c + 0.1 + a \tag{3}\label{eq3}$$

Next, let $t = -0.5$ to get

$$\left| f(-0.5) - g(-0.5) + 2\epsilon \right| = \left| -0.5b - c - 0.5 - a + 2b \right| = \left| 1.5b - c - 0.5 - a \right| \tag{4}\label{eq4}$$

There are many values for $a, b \text{ and } c$ which show that the base requested inequality of

$$|f(x)-g(y)| < |f(t)-g(t)+2\epsilon| \tag{5}\label{eq5}$$

does not hold. For example, if $b = 2$, $c = 1$ and $a = 1$, then \eqref{eq3} gives $3.9$ while \eqref{eq4} give $0.5$.


We have $$ |f(x)-g(x)|= |f(x)-f(t)+f(t)-g(x)+g(t)-g(t)| $$ And this is less than $$ |f(t)-g(t)+2\epsilon| $$ by uniform continuity

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    $\begingroup$ Can you explain how you keep the $\varepsilon$ inside the absolute values? I'm in agreement with this method if you know $f(t)\geq g(t)$ for all $t\in I$, but it's possible that the inequality reverses any number of times. Essentially, one can easily prove $|f(x)-g(y)|<|f(t)-g(t)|+2\varepsilon$ for all $t\in I$, but keeping the $\varepsilon$ inside the absolute values seemed a little more delicate (I can be missing something obvious, though). $\endgroup$
    – Clayton
    Mar 6, 2019 at 22:17
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    $\begingroup$ @Clayton good point...actually doesn't that lead to a counter example showing this is false without additional assumptions? For instance take $f=x$ and $g=x+1$? $\endgroup$
    – TomGrubb
    Mar 6, 2019 at 22:29
  • $\begingroup$ @Clayton I agree with what you wrote, including that it's quite easy to prove that $|f(x)-g(y)| \lt |f(t)-g(t)| + 2\epsilon$ for all $t \in I$. As I believe I show how to create various fairly simple counter-examples in my answer, you can't always keep the $\epsilon$ within the absolute values, so I don't think you're missing anything obvious. $\endgroup$ Mar 7, 2019 at 3:20

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