Let $f\colon [a,b] \rightarrow \mathbb{R}$ be a continuous function. Prove that the function $g$, defined by $g(x)=\max \{f(t); a \leq t \leq x \}$ is uniformly continuous.

My attempt at a proof:

For $x \geq b $ $g$ is obviously constant and for $x<a$ it isn't defined. Let's first prove that $g$ is continuous on [a,b]. $$\forall x \in [a,b]:\lim_{h \searrow 0} g(x+h)=\lim_{h \nearrow 0} g(x+h)$$

$\lim_{h \searrow 0}\max \{f(t); a \leq t \leq x+h \}=\lim_{h \nearrow 0} \max \{ f(t); a \leq t \leq x \}$ because f is continuous on $[a,b]$. The same applies for the other limit, therefore the limits are the same. So $g$ is continuous on $[a,b]$.

Recall the definition of uniform continuity: $$\forall \varepsilon>0 \exists \delta>0 \forall x_1, x_2 \in D_g:(|x_1-x_2|<\delta \Rightarrow |g(x_1)-g(x_2)|<\varepsilon)$$

Because $g$ is continuous on $[a,b]$, it follows that $g$ is uniformly continuous on $[a,b]$. So if $x_1, x_2 \in [a,b] \Rightarrow (|x_1-x_2|< \delta \Rightarrow |g(x_1)-g(x_2)|< \varepsilon)$. The only other options are $x_1 \in [a,b] \wedge x_2 \in (b, \infty) $ or $x_1 \in (b, \infty) \wedge x_2 \in [a,b]$. Because the proof is analogous we will only prove the 1st case.

Let $x_1 \in [a,b] \wedge x_2 \in (b, \infty)$. We prove: $$\forall \varepsilon>0 \exists \delta>0 \forall x_1 \in [a,b] \forall x_2 \in (b, \infty): (|x_1-x_2|< \delta \Rightarrow |g(x_1)-g(x_2)|< \varepsilon)$$

Let $\varepsilon>0$ be arbitrary. Recall that $g$ is constant on $[b, \infty)$. Therefore: $$|g(x_1)-g(x_2)|=|g(x_1)-g(b)|$$

Therefore for $\delta$ we can choose the $\delta$, which assures uniform continuity on $[a,b]$, because $|g(x_1)-g(x_2)|< \varepsilon \iff |g(x_1)-g(b)|< \varepsilon$. Therefore $g$ is uniformly continuous where it's defined.

This concludes the proof.

Is this proof valid?

  • 3
    $\begingroup$ Without reading your proof, what comes to mind is the following: there's the general fact that a continuous, real-valued function on a compact set (of $\mathbb{R}^n$) is always uniformly continuous. So, all you need to show is that $g$ is continuous. $\endgroup$ – Louis Aug 7 '13 at 18:47
  • $\begingroup$ Yes, I used that fact in the proof. Though we must also note that g is defined on the entire real line right of a. $\endgroup$ – gndz Aug 7 '13 at 18:59
  • $\begingroup$ I actually forgot another possibility: x1 and x2 are both on (b, infinity), but that's pretty trivial to prove because g is constant on that interval. $\endgroup$ – gndz Aug 7 '13 at 19:10

Extend $f(x) = f(b)$ for $x > b$. Then $f$ is uniformly continuous.

It can be easily seen that $$ |g(x) - g(y)| ≤ \max\{|f(s) - f(t)|: s, t ∈ [x, y]\} ≤ \max\{|f(s) - f(t)|: |s - t| ≤ |x - y|\} $$ So given $ε > 0$ there is $δ > 0$ such that $|s - t| < δ \implies |f(s) - f(t)| < ε$ since $f$ is uniformly continuous. So using the same $δ$ we obtain $$ |g(x) - g(y)| ≤ \max\{|f(s) - f(t)|: |s - t| ≤ |x - y| < δ\} < ε $$ so $g$ is uniformly continuous.


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