if $f: \mathbb{R}\to \mathbb{R}$ satisfies $$|f(x)-f(y)|\le |x-y|^{\sqrt{2}}$$ for all $x,y\in \mathbb{R}$ ,then is f increasing ,decreasing or constant? in my view ,it is clear that $|f(x)-f(y)|$ is not zero so it is not constant. And since the rate of change is positive so the function would be increasing.Is my solution correct or wrong?
7080
I want to show that if a function $f:[a,b]\rightarrow \mathbb R$ satisfies a Hölder condition of order $\alpha &gt; 1 $ then it is constant.  The way I think of it is as follows:  $$|f(x) - f(y)| &lt; K|x-y|^\alpha$$  $$\frac{|f(x) - f(y)|} {|x-y]}  &lt; K|x-y|^{\alpha -1}$$  $$\lim_{y\rightarrow x} \frac{|f(x) - f(y)|} {|x-y]} \le \lim_{y\rightarrow x}  K|x-y|^{\alpha -1} =0 $$ As the limit is $0$, we can remove the modulus, so we get:  $$\lim_{y\rightarrow x} \frac{f(x) - f(y)} {x-y} = 0$$ So $f$ is derivable and $f'(x) = 0$ for all $x$ in $[a,b]$. Note that the reason we can add the limit $y\rightarrow x$ is because $[a,b]$ is closed in $\mathbb R$.  However, the question gives as a hint using the mean value theorem. I am not sure why one should do that. You would first have to prove that $f$ is derivable in a similar manner to what I did, and then prove that $f$ is constant. Or is there a simpler way to do it and I am missing it?  Also please inform me of any mistakes I did in the proof (if any)/ Thank you!
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
23306
Let $f : \mathbb{R} \to \mathbb{R}$ be a function which is differentiable everywhere on $\mathbb{R}$. If we know that  $$\lim_{x\to-\infty}f(x) = L = \lim_{x\to\infty}f(x)$$ How would one go about showing that there must exist $c \in \mathbb{R}$, such that $f'(c) = 0$ for all cases at once?
4826
I need to show that the following function is constant: $f:(0,\infty)\to\mathbb{R}$, $f(2^x)=f(3^x)$ and $f$ is continuous. I need to find the value of $f(x)$.    So I did something but I do not know how to continue. Let $x\to \log_2 x$ and we get that $f(x)=f(3^{\log_2 x)}$. Now, let $x\to 3^{\log_2 x}$ and we get that $f(3^{\log_2 x)}=f(3^{\log_2 3^{\log_2 x})}$. And so on. I think I need to find a rule. Do you have any idea?
54805
Let $F:\mathbb{R^2}\rightarrow \mathbb{R}$ be a function, such as : $$F(x,y)=\begin{cases}        0 &amp; (x,y)=(0,0) \\       \frac{xy}{x^2+y^2} &amp; (x,y)\neq (0,0)    \end{cases}$$      Is $F$ differentiable at $(0,0)$?   My attempt :  Consider the limit, $\lim_{(x,y)\rightarrow (0,0)} F(x,y)$ along the line $y=x$. The limit equals to $\frac{1}{2}$. However, $F(0,0)=0$. Hence $F$ is not continuous at $(0,0)$ which implies that $F$ isn't differentiable as well.  Is this approach okay ?
101313
Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. $x\in\mathbb{R}$ is a fixed point of $f$ if $f(x)=x$. Show that if $f'(t)\neq 1\;\forall\;t\in\mathbb{R}$, then $f$ has at most one fixed point.  My biggest problem  with this is that it doesn't seem to be true. For example, consider $f(x)=x^2$. Then certainly $f(0)=0$ and $f(1)=1 \Rightarrow 0$ and $1$ are fixed points. But $f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}$. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition.
92984
Given the definition of an increasing function as a function $f:E\to\mathbb{R}$, where $E$ is nonempty such that $x_1,x_2\in E$ and $x_1&lt;x_2$ imply $f(x_1)\leq f(x_2)$, then couldn't, for example, $f(x)=-1$ considered increasing? Since the inequality is not a strict inequality.
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?
83348
Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ be a function. In order to prove that every function that verifies $\lvert f(x)-f(y)\rvert \leq(x-y)^2$ for all $x,y$, is constant, I have considered on applying the Lagrange's theorem. There exists some $c\in[x,y]$ for which:  $$\frac{\lvert f(x)-f(y)\rvert}{x-y}=f'(c)\leq (x-y)$$  Which would certainly lead to success if I'm able to prove that that is $0$, the problem is I don't know how to proceed with the proof. Any help appreciated.
122614
$f$ is an entire function, and it satisfies $f(\mathbb{C}) \subseteq \{z \in \mathbb{C} \mid \operatorname{Im} z &gt; 0\}$. Show that $f$ is constant.   I want to take advantage of the Liouville's Theorem, but I just can't figure out the relationship between its image part with its module.
12440
I'm trying to make a problem for my advanced calculus students.  I was thinking, if we have a differentiable function $f:\mathbb{R}\to\mathbb{R}$ such that $f'(q)=0$ for all $q\in\mathbb{Q}$, can we say that $f$ is constant?
98516
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and differentiable function in all $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all  $x\in\mathbb{R}$, then $f\equiv 0$.  I've been trying to prove this using the Mean Value Theorem, but I can't get to the result. Can someone help?
21582
Suppose $f$ is a differentiable function over $\mathbb{R}$ satisfying $f(0) = 0$. Show that if $f'$ is strictly increasing, then $\frac{f(x)}{x}$ is increasing over $(0,\infty)$.   Attempt:  Since $f'$ is increasing we know that $f'' &gt; 0$ for all $x$. Now let $g(x) = \frac{f(x)}{x}$ and thus $f(x) = xg(x)$. Then we wish to show that $g'(x) &gt; 0$ on $(0,\infty)$ and so since $f'(x) = xg'(x)+g(x)$ and $f''(x) = xg''(x)+2g'(x) &gt; 0$, we have $g''(x) &gt; \frac{-2g'(x)}{x}$. I am not sure how to use this though to prove the statement.
99902
Let $f$ be a continously differentiable function on $\mathbb R$.  Suppose that   $$L=\lim_{x\to\infty}(f(x)+f'(x))$$ exists.  If $0&lt;L&lt;\infty$, then which of the following statements is\are correct?               If $\lim_{x\to\infty} f'(x)$ exists, then it is $0$.     If $\lim_{x\to\infty} f(x)$ exists, then it is $L$.     If $\lim_{x\to\infty} f'(x)$ exists, then $\lim_{x\to\infty}f(x)=0$.     If $\lim_{x\to\infty} f(x)$ exists, then $\lim_{x\to\infty}f'(x)=0$.           My Guess  I could not conclude the answer and prove that properly. But, I guess that it must be 1 and 2. help me.
81123
Let $f : \mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function such as $\forall x \in \mathbb{R}, f'(x) &gt; 2$. What could I do to show that $\exists c \in \mathbb{R} : f(c) = 0$ ? At first I thought I could use the mean value theorem but it did not help me. In fact, I don't really understand why $f(x)$ couldn't be strictly positive $\forall x \in \mathbb{R}$. Would you know how to proceed ? Thanks for your help.
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 :)
9106
This is a homework problem for my real analysis class. It is problem 13.1 from Mattuck's Introdution to Analysis. The question reads:       Suppose F(x) is continuous on some open interval I and c is a maximum point inside this interval. Is it true that f(x) must be increasing immediately to the left of c and decreasing immediately to the right of c? Proof or counterexample. (Note: A constant function is considered to be both increasing and decreasing.)      My friend and I have been attempting to find a counterexample. We figured a function that begins oscillating infinitely many times near c but converges to c would work. We're having difficulty constructing such a function out of elementary functions so we think we will have to define it in another way, piece-wise using receding intervals. I have no clue how this would look. Any advice would be greatly appreciated.