query: continuous function $h(x)=\sup\{f(t)\colon t\le x\}$
positive_passages.docid: 144614
positive_passages.text: If $f$ is continuous on $[a,b]$ and $F(x) = \sup f([a,x])$. Prove that $F$ is continuous on $[a,b]$ .
negative_passages.docid: 62048
negative_passages.text: Proving a Lipschitz function is continuous
negative_passages.docid: 131992
negative_passages.text: A problem on continuous functions
negative_passages.docid: 87239
negative_passages.text: Prove functions defined by sup and inf are continuous
negative_passages.docid: 75735
negative_passages.text: Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(\frac{1}{i})$ (for large enough $i$)
negative_passages.docid: 88586
negative_passages.text: show that every continuous real-valued function defined on $S_{\mathbb{\Omega}}$ is eventually constant
negative_passages.docid: 111990
negative_passages.text: Continuous function from $(0, 1]$ onto $(0, 1)$?
negative_passages.docid: 2142
negative_passages.text: Finding a set of continuous functions with a certain property 2
negative_passages.docid: 28730
negative_passages.text: product of two uniformly continuous functions is uniformly continuous
negative_passages.docid: 111485
negative_passages.text: Let $f(x)$ be continuous on [$0,1$] to $f(x)=0$ for each $x\in[0,1]$
negative_passages.docid: 41690
negative_passages.text: Products of Continuous functions: Topology
negative_passages.docid: 118332
negative_passages.text: proof $f(\overline{A})\subseteq \overline{f(A)} \Leftrightarrow f$ continuous
negative_passages.docid: 136011
negative_passages.text: Must $f$ be continuous?
negative_passages.docid: 34836
negative_passages.text: Composition of Absolutely Continuous Functions
negative_passages.docid: 158567
negative_passages.text: A uniform continuous function which is not Hölder continuous
negative_passages.docid: 104510
negative_passages.text: For continuous $f$, $\int_0^x f(t)dt = ∫_x^1f(t)dt$ implies $f=0$.
negative_passages.docid: 103495
negative_passages.text: Let $f$ be continuous. Then $\int_{0}^{x}\left[\int_{0}^{t}f\left(u\right)du\right]dt=\int_{0}^{x}f\left(u\right)\left(x-u\right)du$