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Exercise:

Suppose that $f$ is continuous on $[a,b]$ and that $F(x) = \sup f([a,x])$. Prove that $F$ is continuous on $[a,b]$ .

Attempt of proof: Suppose that $f$ is continuous on $[a,b]$ and that $F(x) = \sup f([a,x])$. Then since $f$ is continuous on [a,b]. Then by definition $f$ is continuous at a point $c$ in $(a,b)$ iff for every $ε > 0$ there is a $δ > 0$ such that $|c-x| < δ$ $→$ $|f(c) - f(x)| < ε$. Then we need to show $F$ is continuous.

Then let $ε > 0$, then there is a $δ>0$ such that $|c-x| < δ$.

Then, $|F(c) - F(x)| = |\sup f([a,c)] - \sup f([a,x)]|$.

Can someone please help me? I am stuck and don't know how to continue. Thank you in advance.

Przemysław Scherwentke
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Mahidevran
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1 Answers1

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Hint: you need to show for any $c$ and $\epsilon$ there exists $\delta$ such that $|x-c|<\delta$ gives $|F(x)-F(c)|<\epsilon$. The continuity of $f$ implies that for any $x_0$, $\epsilon$ there exists $\delta$ such that $|x-x_0|<\delta$ gives $|f(x)-f(x_0)|<\epsilon$.

Case 1: $\sup f([a,c]) > f(c) $. Then by continuity you know there exists $\delta_0$ such that for any $x$ such that $|x-c|<\delta_0$ you get $f(x)<\sup f([a,c])$.

Case 2: $\sup f([a,c]) = f(c) $. Again by continuity you for any $\epsilon$ there exists $\delta$ such that $|x-c|<\delta$ gives $f(x)<f(c)+\epsilon$.

Plug the 2 cases into $|F(x)-F(c)| = |\sup f([a,x])-\sup f([a,c])|$ allows to conclude...

Milly
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  • in both cases only $supf([a,c])$ is considered. We don't need $supf([a,x)]$? – Mahidevran Nov 10 '14 at 00:58
  • $\sup f([a,c])$ "controls" $\sup f([a,x])$ in both cases (for $|x-c|<\min{\delta_0,\delta}$). Combining the 2 cases you get that for any $c$, $\epsilon$ there exists $\delta$ such that $|x-c|<\delta$ gives $|\sup([a,c])-\sup(f[a,x])|<\epsilon$. – Milly Nov 10 '14 at 01:03