Let $f$ a continuous function on all $\mathbb R$. How can I prove that if $f\circ f\circ f=id$, then $f=id$ ? I really have no idea.
122796
Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function such that $f^3(x)=f\circ f\circ f(x)=x$ for all $x$.  How can I prove $f$ is the identity function?
39699
Greets   I came up the other day with the following question: Is it true that $f:\mathbb{R}^n\longrightarrow{\mathbb{R}^m}$ is continuous if and only if $f$ maps compact sets onto compact sets and maps connected sets onto connected sets?  I'm having trouble showing the "if", and I haven't found a counterexample to this part, so I would appreciate ideas to prove or disprove this.  Thanks
108878
Let $\mathbb S^1$ denote the unit circle in $\mathbb R^2$.      Then prove that for every  continuous function $f:\mathbb S^1 \to \mathbb R$, there exist uncountably many pairs of distinct points $x, y$ in $S^1$, such that $f(x)=f(y)$.
10735
I am stuck with the following problem that says :     let $f \colon [a,b] \to \mathbb{R}$ be continuous on $[a,b]$    and $f''$ exist for all $x \in (a,b)$. Let $a \lt c \lt b$.    Prove that there exists a point $\xi \in (a,b)$    such that $$f(c)=\frac{b-c}{b-a}f(a)+\left( \frac{c-a}{b - a} \right)f(b)+\frac12(c-a)(c-b)f''(\xi).$$   Can someone explain in details? Thanks in advance for your time.
76724
I have proved that if $f \in R[a,b]$ and given $\epsilon &gt; 0$ there exists a continuous function $g$ such that $$\int_a^b |f-g| &lt; \epsilon$$ I was wondering if using this fact there is some way to show that there is also some continuous function $h$ such that $$\int_a^b |f-h|^2 &lt; \epsilon$$ Any help will be appreciated, thanks :)
19321
I am thinking about this since couple hour. Is a continuous and bounded function $f:\mathbb R\to\mathbb R$ uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let $\epsilon&gt;0$ and $a\in\mathbb R$. By continuity, there is a $\delta_a&gt;0$ such that $$d(f(x),f(y))&lt;d(f(x),f(a))+d(f(y),f(a))&lt;\epsilon$$ if $d(x,y)\leq d(x,a)+d(y,a)&lt;\delta_a.$ Then, if $\inf_a\delta_a&gt;0$ I set $\delta=\inf_a\delta_a$ and we are done since $d(f(x),f(y))&lt;\epsilon$ if $d(x,y)&lt;\delta,$ but I didn't success to prove that $\delta\neq 0$.
92142
We have the function $f: \mathbb{R} \to \mathbb{R}$.     Question: Show that if the sets $\{x:f(x) &gt; a\}$ and $\{x:f(x) &lt; a\}$ are open for every $a\in\mathbb{R}$, then $f$ is continuous.   My attempt:$f$ is continuous if $A \subset \mathbb{R}$ is open, then $f^{-1}(A) \subset \mathbb{R}$ is open. I can show that the image of $\{x:f(x) &gt;a\}$ is open. Namely $\{f(x) &gt; a\} = (a,\infty)$ is open in $\mathbb{R}$. But this doesn't prove that for any open set $A \subset \mathbb{R}$, we have that $f^{-1}(A)\subset\mathbb{R}$ open as well.    Thanks in advance!
47071
Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.  WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).  I build up this question while thinking of the question .
2015
Suppose that $f$ is continuous at $x_0$ and $f$ satisfies $f(x)+f(y)=f(x+y)$. Then how can we prove that $f$ is continuous at $x$ for all $x$? I seems to have problem doing anything with it. Thanks in advance.
68189
Prove that if $f\colon\mathbb{R}\to\mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ for all $x,y$, and $f$ is continuous at $0$, then it is continuous everywhere.   If there exists $c \in \mathbb{R}$ such that $f(c) = 0$, then $$f(x + c) = f(x)f(c) = 0.$$ As every real number $y$ can be written as $y = x + c$ for some real $x$, this function is either everywhere zero or nowhere zero. The latter case is the interesting one. So let's consider the case that $f$ is not the constant function $f = 0$.  To prove continuity in this case, note that for any $x \in \mathbb{R}$ $$f(x) = f(x + 0) = f(x)f(0) \implies f(0) = 1.$$  Continuity at $0$ tells us that given any $\varepsilon_0 &gt; 0$, we can find $\delta_0 &gt; 0$ such that $|x| &lt; \delta_0$ implies $$|f(x) - 1| &lt; \varepsilon_0.$$  Okay, so let $c \in \mathbb{R}$ be fixed arbitrarily (recall that $f(c)$ is nonzero). Let $\varepsilon &gt; 0$. By continuity of $f$ at $0$, we can choose $\delta &gt; 0$ such that $$|x - c| &lt; \delta\implies |f(x - c) - 1| &lt; \frac{\varepsilon}{|f(c)|}.$$  Now notice that for all $x$ such that $|x - c| &lt; \delta$, we have $$\begin{align*} |f(x) - f(c)| &amp;= |f(x - c + c) - f(c)|\\  &amp;= |f(x - c)f(c) - f(c)|\\  &amp;= |f(c)| |f(x - c) - 1|\\  &amp;\lt |f(c)| \frac{\varepsilon}{|f(c)|}\\   &amp;= \varepsilon. \end{align*}$$ Hence $f$ is continuous at $c$. Since $c$ was arbitrary, $f$ is continuous on all of $\mathbb{R}$.  Is my procedure correct?
12516
Let $f: X \to Y$ be a function. The graph of $f$ is defined to be the set $G = \{(x, f(x)) : x \in X\}$. Prove that if $f$ is continuous then $G$ is homeomorphic to $X$.     an anyone suggest  me the process of solving this problem, thanks.
63687
I've been given a proof that shows the following:  If $f:[a,b]\to \mathbb C$ is a continuous function and $f(t)=u(t)+iv(t)$ then $$\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$$ The proof begins by letting $\theta$ be the principle argument of the complex number $\int_a^b f(t)dt$ and there is one step in the proof I don't understand; can anyone explain to me why $$\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)$$  Would this not imply that the imaginary part of the left hand side is equal to $0$? I'm not sure why the LHS and RHS would be equal here.
105869
I am supposed to prove that if two continuous functions $f:\mathbb R\rightarrow\mathbb R$ and $g:\mathbb R\rightarrow\mathbb R$ coincide on a dense subset $\\D$ of $\mathbb R$, then they coincide everywhere.  I know that since $\\f$ and $\\g$ are continuous at every point in $\mathbb R$, then the preimage of any open neighborhood of any real number $\\a$ must also be an open neighborhood of $\\f^{-1}(a)$.  But I'm lost on how to use that to get to the conclusion.  Do I have to use the concept of connected sets?
63505
If $F: \mathbb R^2 \to \mathbb R$ and $F_x$ (partial derivative of $F$ wrt $x$) and $F_y$ exist at $(x_0,y_0)$ then the function is continuous at that point. Is this true? If not what could be a counter-example?
38024
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 $ε &gt; 0$ there is a $δ &gt; 0$ such that $|c-x| &lt; δ$  $→$ $|f(c) - f(x)| &lt; ε$.  Then we need to show $F$ is continuous.   Then let $ε &gt; 0$, then there is a $δ&gt;0$ such that $|c-x| &lt; δ$.   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.
42395
If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
77224
Let $f:\mathbb{R}\to\mathbb{R}$ , there is $k \in \mathbb{R}$, $|k|\ne 1$, so that for every $x\in \mathbb{R} , f(kx)=f(x).$ Show that if $f$ continuous at $0$ then $f$ is a constant function.   I tried to find something to show it from the definition of continuous function but I didn't know how to continue. We have that $f$ continuous at $0$ if for every $\epsilon&gt;0$ there is $\delta&gt;0$ such that $|x-0|&lt;\delta \implies |f(x)-f(0)|&lt;\epsilon$ And if $f$ not continuous on $0$ is $f$ still a constant function?