Let $ f : \mathbb R \to \mathbb R $ be such that $ f ( x + y ) = f ( x ) f ( y ) $ for all $ x , y \in \mathbb R $. If $ f $ is continuous at $ x = 0 $ then show that $ f $ is continuous on $ \mathbb R $. Also show that there exists a constant $ c \in \mathbb R $ such that $ f ( x ) = e ^ { c x } $ for all $ x \in \mathbb R $.  Anyone please help me with the solution.
22165
The  asks about functions $f \colon \mathbb R \to \mathbb R$ such that $$f(x+y)=f(x)+f(y).$$ It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called . Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.) Is there some overview of basic facts about Cauchy equation and related functional equations - preferably available online?
110092
I need help to prove these theorems: 1)If $\mathcal f $ is continuous on $\mathbb C $  and $\mathcal f(z) = f(2z)  \forall z \in \mathbb C $ then $\mathcal f $ is constant .  2)If $\mathcal f $  is entire and $\mathcal f(2z)=2f(z)$ then $ \exists $ $\lambda $ such that $\mathcal f(z)=\lambda\mathcal z $.
42627
Let  $ f: \Bbb R^2 \to \Bbb R $ such that :  $ \forall _{y_0 \in \Bbb R\ }:  $ function $ x \to f(x,y_0) $ continuous function and increasing  $ \forall _{x_0 \in \Bbb R\ }:  $ function $ y \to f(x_0,y) $ continuous function   I mean the continuity of one variable.  Prove the continuity of a function $f: \Bbb R^2 \to \Bbb R $  I know definition, but I can not do. I wanted to use the definitions of Cauchy.
21472
Let $f:[0,1] \to \mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x \in [0,1]$.  Any hint/idea for proving that $f$ has to be constant?
87532
Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.  No clue here. Can you please help?
67821
Assume  $f : I \rightarrow \mathbb{R}$ is a non-decreasing on an open interval $I$ and that $f$ satisfies the Intermediate value property or  on $I$ (that is, for any $a &lt; b$ in $I$ and any $L$ between $f(a)$ and $f(b)$, there exists $c \in [a, b]$ such that $ f (c) = L)$. Then, prove that $f$ is continuous.  However, I know that a function can be discontinuous and also satisfy the IVT at the same time. Could someone point me in the right direction?
114828
Let $f: \mathbb{R} \to D \subseteq \mathbb{R}$ be a continuous function. Is there a function $f$ that satisfies the following property? $\forall y \in D$, there are exactly 3 $ x_1,x_2,x_3 \in \mathbb{R}$ such that $f(x_1) = f(x_2) = f(x_3) = y$.
106948
Prove that if $f, g$ : $X$ → $\mathbb R$ are continuous at $a$ ∈ $\mathbb R$, then $f · g$ is continuous in $a$.  If $f$ is continuous at $a$, then $∀ε_f &gt; 0, ∃δ_f &gt; 0$ such that  $|x-a| &lt; δ_f$ iff $|f(x) - f(a)| &lt; ε_f$  and if $g$ is continuous at $a$, then $∀ε_g &gt; 0, ∃δ_g &gt; 0$ such that  $|x-a| &lt; δ_g$ iff $|g(x) - g(a)| &lt; ε_g$.  Now make $|f(x)g(x) - f(a)g(a)|$ =  $|f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)|$ $≤$  $|f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)|$ &lt; $ε_f|g(a)| + ε_g|f(x)|$. (1)  Notice that if $|f(x) - f(a)| &lt; ε_f$ then $|f(x)| - |f(a)| &lt; ε_f$, so $|f(x)| &lt; ε_f + |f(a)|$.  That implies, from (1): $|f(x)g(x) - f(a)g(a)| &lt; ε_f|g(a)| + ε_g(ε_f + |f(a)|) = ε_f(|g(a)| + ε_g|f(a)|)$.  Now I'm a little lost. From the proofs I've seen, it seems to me that I could simply take $δ = min(δ_g, δ_f)$. Also, since $ε_f$ and $ε_g$ can be made as small as one wants, there will be some δ that satisfies the conditions for continuity. But from what I read, I should explicitly show a δ in terms of ε. Anyway, I reasoned that since $f$ and $g$ are continuous, I can make bounds on $x$ symmetrically* as done to $f(x)g(x)$ and got the following:  δ = $δ_f(|a| + δ_g|a|)$ = $δ_f|a|(1 + δ_g)$  so:  $|x-a| &lt; δ_f|a|(1 + δ_g)$ iff $|f(x)g(x) - f(a)g(a)| &lt;  ε_f(|g(a)| + ε_g|f(a)|)$   I need to clarify/better define this, but it is basically the notion that making the same operations I made on the bounds of $f(x)g(x)$ in relation to the bounds of $f(x)$ and $g(x)$ individually over the bounds of $x$ for each function, should leave me with the adequate bounds for $x$ on the composite function.
41723
Let $x_0 \in \mathbb{R}$ and $f,g: \mathbb{R} \to \mathbb{R}$  such that $f$ is discontinuous at $x_0$ but $g$ is continuous at $x_0$, then $f+g$ at $x_0$ is...   My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is $$f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 $$  Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying.   Assume $f+g$ is continuous at $x_0$ that is $$\forall \epsilon &gt; 0 ,\exists \delta_1 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| &lt; \epsilon  $$  Since this statement must be true for all $\epsilon &gt; 0$ I could choose $\epsilon:=1$. Next I know that also $g$ is continuous at said point $x_0$, thus  $$\forall \epsilon &gt; 0, \exists \delta_2 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_2 \implies |g(x)-g(x_0)| &lt; \epsilon $$  and I have that $f$ is NOT continuous at $x_0$ therefore: $$\exists \epsilon^* &gt; 0, \forall \delta&gt;0: \exists x \in  \mathbb{R}, |x-x_0| &lt; \delta \wedge |f(x)-f(x_0)| \geq \epsilon $$  My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose $\epsilon &gt; 0$ such that the last statement is met, so say $\epsilon:= \epsilon^*&gt;0$, so statement 1 and 2 are still true for said $\epsilon$  I also need to choose $\delta$. Lets say $\delta = \min{(\delta_1,\delta_2)}$ now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation.  Let $|x-x_0|&lt; \delta$, then we obtain $$ |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| &lt; - \epsilon + \epsilon =0 $$ which means that $|f(x)+g(x)-f(x_0)-g(x_0)| &lt;0$ and I hope that this is a contradiction because $|.| \geq 0$. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation.
53589
If $f$ is continous in $[a,b] \subset \mathbb{R}$ and exist real numbers $\alpha \not= \beta$ such that $$\alpha\int_{a}^{c} f(x) dx + \beta\int_{c}^{b} f(x) dx  =0$$ for all $c \in [a,b]$ show that $f (x)=0$ for all $x \in [a,b].$
119903
Whenever $X$ is a topological space, let us say that continuity on $X$ is detected by $\mathbb{R}$ iff for all functions $f : X \rightarrow Y$ where $Y$ is another topological space, we have that if every continuous function $c : \mathbb{R} \rightarrow X$ satisfies "$f \circ c$ is continuous", then $f$ is itself continuous.   Clearly, continuity on $\mathbb{R}$ is detected by $\mathbb{R}$ It is shown  that continuity on $\mathbb{R}^2$ is detected by $\mathbb{R}$. Clearly, it is not the case that continuity on $\mathbb{Q}$ is detected by $\mathbb{R}$, since every continuous mapping $\mathbb{R} \rightarrow \mathbb{Q}$ is constant.      Question. Do there exist general conditions underwhich we can conclude that continuity on a topological space is detected by $\mathbb{R}$?
2291
Question is to prove :  $f : S^1 \to \mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$  I guess it would be helpful to use intermediate value theorem  Assuming $f(x)\neq f(-x)$ then given any $p\in (f(-x),f(x))$ (assuming $f(-x)&lt;f(x)$) there exists $y\in (-x, x)$ such that $f(y)=p$  I am not very sure of how to use this..   It would be helpful if someone can give some hint which would help me to solve this..  Thank you.
37702
$(i)$ Given a function $f:E \rightarrow \mathbb{R}$ define $|f|:E \rightarrow \mathbb{R}$ by $|f|(x)=|f(x)|$ for $x \in E$ show if $f$ is continuous at $c \in E$ then so is $|f|$.        $(ii)$ Now the converse is not true, find $f$ such that $|f|$ is continuous but $f$ is not.
89560
I am having difficulties showing that this is true. How would you prove this statement?  Given that the function $h: \mathbb{R}\rightarrow\mathbb{R}$ is increasing. Show that $h$ is continuous if and only if  $h(\mathbb{R})$ is connected.
3532
Does there exist a continuous function $\: f : \mathbf{R} \to \mathbf{R} \:$ such that for   all  functions $\: g : \mathbf{R} \to \mathbf{R} \:$, for all real numbers $x$,  there exists a real number $y$ such that $\: x &lt; y \:$ and $\: g(y) &lt; f(y) \:$?
107447
Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.   $g(0)=1$. If $a=g(1)$,then $a&gt;0$ and $g(r)&gt;a^r$ for all $r$ in $\mathbb Q$.   Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.
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!