Show that there is a continuous function $g:\Bbb R\to \Bbb R$ with $g(x)=f(x)$ for every $x\in C$.
128911
Continuous extension of a real function
122160
$f:\mathbb R \to \mathbb R$ be a continuous such that $f(f(f(x)))=x,\forall x \in \mathbb R$ , is $f(x)=x , \forall x \in \mathbb R$?
77267
Given $f$ is continuous and $f(x)=f(e^{t}x)$ for all $x\in\mathbb{R}$ and $t\ge0$, show that $f$ is constant function
73594
Existence of continuous function $f$ on $\Bbb R$ which vanishes exactly on $A\subset \Bbb R$
62723
If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere
113687
$f$ is continuous function from $ [0,1]$ to itself and f(0)=0, f(1)=1 and $f(f(x))=x$ then show that $f(x)=x$ on$[0,1]$
14495
I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$
58597
A characterization of functions from $\mathbb R^n$ to $\mathbb R^m$ which are continuous
121708
$S := \{x \in \Bbb R^3: ||x||_2 = 1 \}$ and $T: S^2 \to \Bbb R$ is a continuous function. Is $T$ injective?
25010
Argue that there exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ which satisfies $g(x) = \frac{1-e^{-2x^2}}{x^2}$ for $x \neq 0$
7876
If $f(x+y)=f(x)+f(y) ,\forall\;x,y\in\Bbb{R}$, then if $f$ is continuous at $0$, then it is continuous on $\Bbb{R}.$
54403
Given $f : \Bbb{R} \to \Bbb{R}$ is a continuous function. To prove that $\int_{0}^{1}f(x)x^2 dx = \frac{f(c)}{3}$ for some $c \in [0,1]$
138105
Let $f$ and $g$ be continuous at $ x=a$, with $f(a)<g(a)$. Prove that there exists a neighborhood $I$ of a such that $f(x)<g(x)$ for all $x \in I$.
19916
$f(x)=f(x^2+ 1/4)$ , $f$ is continuous from $\mathbb{R}$ to $\mathbb{R}$
68165
Find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous at precisely one point?
86472
$f:\mathbb R \to \mathbb R^n$ be such that $G(f):=\{(x,f(x)):x \in \mathbb R\}$ is closed and connected in $\mathbb R^{n+1}$ , is $f$ continuous ?
90565
If $f:[0,1] \rightarrow \mathbb{R}$ is continuous function such that $f(0)=f(1)$ then there exists $ x\in [0,1]$ such that $f(x) = f(x+\frac{1}{n})$.