Let $f: \mathbb{R} \to \mathbb{R}$ be continuous at $0$ and $f(x+y) = f(x) + f(y)$ for all $x, y$ in $\mathbb{R}$. Show that there is a real number $c$ such that $f(x) = cx$.  I'm not sure about how to do show this, I can tell that $f(0) = 0$ and that from continuity at $0$ we get $|f(x)| &lt; \epsilon$ for $x$ in some delta neighborhood of $0$. Can anyone help me with this please. Edit: please note that the function is only continuous at $0$ so Cauchy functional equation properties can't be used here.
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?
65616
Let $f$ be a linear functional on a normed space $(X, \|\cdot\|)$. Prove that $f$ is continuous at $x_0=0$ if and only if $f$ is continuous at every $x\in X$.  I understand that the $\Leftarrow$ is trivial but what about the other way?
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.
87540
This question was asked in :     Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that   for all $x\in\mathbb{R}$ and $t\ge 0$, $$f(x)=f(e^{t}x)$$ Show that   $f$ is a constant function.     My attempt:  Suppose that $f$ is not a constant function. Then $f(0)\ne f(x_0)$ for some $x_0 \in \mathbb{R}$. We eliminate the possibilities that $x_0&gt;0$ and $x_0&lt;0$, thus proving that our assumption was wrong.  Case 1: ($x_0&gt;0$). Let $k$ be any real number between $f(0)$ and $f(x_0)$ (not inclusive). Then by the intermediate value theorem,  there exists $y_0 \in (0, x_0)$ such that $f(y_0)=k$. But $f(y_0)=f\left( e^{\ln \left( \frac{x_0}{y_0}\right)  } y_0\right) = f(x_0)$ which contradicts our assumption that $f(y_0)$ was between  $f(0)$ and $f(x_0)$.  Case 2: ($x_0&lt;0$). Let $k$ be any real number between $f(0)$ and $f(x_0)$ (not inclusive). Then by the intermediate value theorem,  there exists $y_0 \in (x_0, 0)$ such that $f(y_0)=k$. But $f(x_0)=f\left( e^{\ln \left( \frac{y_0}{x_0}\right)  } x_0\right) = f(y_0)$, a contradiction again.    Is this proof correct? I was probably looking for a direct proof if there's any. Alternative proofs are welcome.
87979
Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.   Please, see if I made some mistake in the proof below. I mention some theorems in the proof:   The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.  (1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.   (2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.   (3) We'll do this by the Precise Limit Definion: given $ \epsilon&gt;0, \exists\delta|0&lt;|x-x_0|&lt;\delta$, then $0&lt;|f(x)-f(x_0)|&lt;\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.
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.
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.
114512
One of my friend gave the following problem to me but I am unable to solve it even after trying it for nearly two days.     Problem. Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=f(1)=0$. If $f$ is differentiable on $(0,1)$ then prove that there exists $c\in(0,1)$ such that $f'(c)=f(c)$.   Can anyone help me solving this problem?
23490
If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.   Is the proof trivial as it seems or does it need an analytical approach?
120179
I'm having trouble with this functional equation, to find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $f(x) + f(x+f(y)) = y + f(f(x) + f(f(y)))$ I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity. So 0 progress. Tried with $x=y=0, x=0, y=0, x=-f(y), y=f(x)$, and even with some assumptions on bijectivity, but nothing helped. I don't know the source but seems like a pretty difficult problem, maybe even impossible to solve.
18361
This is part of a homework assignment for a real analysis course taught out of "Baby Rudin."  Just looking for a push in the right direction, not a full-blown solution.  We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero.  The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant.  I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers.  Is this an approach worth taking?
11660
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance.
78312
Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
117277
I know that this question has been asked here before but I want to use a different approach. Here is the question.  A function $f:\Bbb{R}\to\Bbb{R}$ is such that  \begin{align} f(x+y)=f(x)+f(y) ,\;\;\forall\;x,y\in\Bbb{R}\qquad\qquad\qquad(1)\end{align} I want to show that if $f$ is continuous at $0$, it is continuous on $\Bbb{R}.$  MY WORK  Since $(1)$ holds for all $x\in \Bbb{R},$ we let \begin{align} x=x-y+y\end{align} Then,  \begin{align} f(x-y+y)=f(x-y)+f(y)\end{align} \begin{align} f(x-y)=f(x)-f(y)\end{align} Let $x_0\in \Bbb{R}, \;\epsilon&gt;$ and $y=x-x_0,\;\;\forall\,x\in\Bbb{R}.$ Then,  \begin{align} f(x-(x-x_0))=f(x)-f(x-x_0)\end{align}  \begin{align} f(x_0)=f(x)-f(x-x_0)\end{align}  \begin{align} f(y)=f(x_0)-f(x)\end{align}  HINTS BY MY PDF:  Let $x_0\in \Bbb{R}, \;\epsilon&gt;$ and $y=x-x_0,\;\;\forall\,x\in\Bbb{R}.$ Then, show that  \begin{align} \left|f(x_0)-f(x)\right|=\left|f(y)-f(0)\right|\end{align} Using this equation and the continuity of $f$ at $0$, establish properly that  \begin{align}\left|f(y)-f(0)\right|&lt;\epsilon,\end{align} in some neighbourhood of $0$.  My problem is how to put this hint together to complete the proof. Please, I need assistance, thanks!
53477
Suppose $\epsilon &gt; 0$ and that $c \in \mathbb{R}$. Since $f^3$ is continuous, we can choose $\delta &gt; 0$ such that whenever $\lvert x - c \rvert &lt; \delta$ we have that $\rvert f^3(x) - f^3(c) \lvert &lt; \epsilon$.   To show that $f$ is continuous, I know I need some number $\delta_1 &gt; 0$ such that $\lvert x - c \rvert &lt; \delta_1$ implies that $\lvert f(x) - f(c) \rvert &lt; \epsilon$.   I've tried manipulating the expression $\lvert f(x) - f(c) \rvert &lt; \epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get  $$   \lvert f(x) - f(c) \rvert &lt; \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) + f^2(c) \rvert} &lt; \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) \rvert}. $$ However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
55720
Do there exist functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y),$ but which aren't linear? I bet you they exist, but I can't think of any examples.  Furthermore, what hypotheses do we need to put on $f$ before no such functions exist? I feel continuity should be enough.
81260
How do I prove that if $f: \Bbb{R}\rightarrow\Bbb{R}$  is continuous at $0$ and $$f(x)=f(2x)$$ for each $x\in\Bbb{R}$, then $f$ is constant?