Visually it seems that a strictly decreasing twice-differentiable map $f: [0, \infty) \to \mathbb{R}$ with $f''(x) &lt; 0$ for all $x$ should be negative at some point, but I can't come up with a formal proof.  Any help?
123459
If $f(x) &gt; 0$ over the reals is it possible to have $f'(x) &lt; 0$ and $f''(x)&lt; 0 $ over the reals? Assuming $f$ can be differentiated twice.
11949
Put precisely, if $f:\mathbb R\to\mathbb R$ is continuous with  $$ \lim_{s\to 0^+}s^{-1}(f(t+s)-f(t-s))=0 \kern20mm (*) $$ for all $t\in\mathbb R$ , is $f$ then constant?  I guess it is but I cannot see the proof. Obviously, if instead of ($*$) we had $\lim_{s\to 0^+}s^{-1}(f(t+s)-f(t))=0$ , i.e. the right-handed derivative zero, then the assertion could be proved indirectly by a simple compactness argument applied to closed intervals $[a,b]$ with $f(a)\not=f(b)$ . However, I cannot see how this argument could be modified to get the above result. Is there possibly a reference to this, or is the result even false?
92929
Assume that $f\colon\mathbb{R\to R}$ is an infinitely often differentiable function, and that a point $a\in\mathbb R$ satisfies $f^{(k)}(a) = 0$ for alle $k\ge 1$. Does there exist an $\epsilon &gt; 0$ such that $f$ is constant at $(a-\epsilon, a+\epsilon)$? If not, can somebody find a counterexample?
80936
$f(x)$ is a continuos fuction in $[0, 1]$ and differentiable twice on $0$.  $U_n=(-1)^{n}f(\frac{1}{n})$. I need to prove that:   $1.$if $\sum_{n=1}^\infty U_n$ converge then $f(0)=0$.  $2.$if $f(0)=f'(0)=0$ then $\sum_{n=1}^\infty U_n$ converge absolutely.  Now, from the detail that $f(x)$ is differentiable twice on $0$ I think I need to write $f(1/n)$ with Maclaurin series with 2 degree but it doesn't work very nice. can you guide me please? thanks!
59414
How would you show that there is no continuous injective map  $f:\mathbb{R}^3 \to \mathbb{R}$?  I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function otherwise $\mathbb{R} \backslash \{a\} $ would be connected?
37274
For a differentiable function $f:\mathbb R\to\mathbb R$ I need to choose the correct statement(s):  $f'(x)\le r&lt;1~\forall~x\in\mathbb R\implies f$ has at least one fixed point.  $f'(x)\le r&lt;1~\forall~x\in\mathbb R\implies f$ has a unique fixed point.    My attempt:  $g:\mathbb R\to\mathbb R:x\mapsto f(x)-x$ is differentiable on $\mathbb R.$ The search is for the number of roots of $g.$ Now $f'(x)\le r&lt;1~\forall~x\in\mathbb R\implies g'(x)\leq r-1&lt;0~\forall~x\in\mathbb R\implies g$ is decreasing everywhere on $\mathbb R.$ Can I say something from this observation$?$
65966
Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})&lt;0$ for every $0&lt;x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0&lt; \delta &lt; 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})&lt;0$ So by Darboux Theorem, there exists a $x_{0}&lt;c&lt; \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)
73790
Let $f: \mathbb R ^n\to\mathbb R$ be a differentiable function.   If $x_0\in \mathbb R^n$ is a point of local minimum of $f$, then $\nabla f(x_0) = 0$.   Where can I find a proof for this theorem? This is a theorem for max/min in calculus of several variables.  @Peter Tamaroff   Here is my attempt:  Let $x_0$ = [x1,x2,...xn]  Let g$_i$(x) = f(x$_0$+(x-x$_i$)e$_i$) where e$_i$ is the ith standard basis vector of dimension n.   Since f has local min at x$_0$, then g$_i$(x) has local min at x$_i$.   So by Fermat's theorem, g'(x$_i$)= 0 which is equal to f$_{x_i}$($x_0)$   Therefore f$_{x_i}$($x_0)$ = 0 which is what you wanted to show. Is this right??? Any comments would be appreciated.
85253
I am working through a bank of previous exams and couldn't figure a problem out to my satisfaction.     Let $f(x) : \mathbb{R} \to \mathbb{R}\,$ be a continuous function.         Show that $f$ can have at most countably many strict local maxima.   Assume that $f$ is not monotone on any interval. Then show that the local   maxima of $f$ are dense in   $\mathbb{R}$.
51130
Let $f:U\rightarrow U$  be analytic, where $U \subsetneqq \mathbb{C}$ is a simply connected domain. Suppose that $f$ has at least two fixed points. Prove that $f$ is the indentity map.  I am able to prove the theorem in the case where $U = \mathbb{D}$ is the unit disc (the proof uses Schwarz lemma, and is discussed ).  The proof however skips the (rigorous) details of how the Riemann Mapping Theorem allows us to reduce from $U$ to $\mathbb{D}$. Any details are appreciated.
122170
Suppose $f(0) =0 $ and  $0&lt;f''(x)&lt;\infty (\forall$  $x&gt;0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases.    I have shown that $f'(x)-\frac{f(x)}{x}  = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here?
93339
Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $,  $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.  $\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq  K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea.
115359
I have a function $f: [0,1]\rightarrow \mathbb{R}$ that is twice differentiable.   For this function is stated that there is a line segment with endpoints $A=(0,f(0))$ and $B=(1,f(1))$ that cuts function $f$ at $(a,f(a))$ where $0&lt;a&lt;1$.   I have to prove that there exists such $x_0$ that $f''(x_0)=0$.   Should I use some theorem for solving this? Like Rolle or Lagrange?  Somehow it seems like I should formulate a function between these two points or something, but I don't know whether it's linear or not.  Any help would be appreciated. Thank you in advance.
24282
This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".     For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.   The proof for the statement in bold is the following:     Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.   I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise.
73846
I've been stuck on this one for a while. Comes from an analysis qual question.  Let f be a function that is continuous on $\left[0,1\right]$ and differentiable on $(0,1)$. Show that if $f(0)=0$ and $|f'(x)| \leq |f(x)|$ for all $x \in (0,1)$, then $f(x)=0$ for all $x \in \left[0,1\right]$.  What I've tried doing so far is see if there was anything I could do with MVT. I didn't really see anything to do with definitions either..to which I have a feeling I'll be playing around with them. Drawing a picture was a little difficult with these conditions as well  Any hints/suggestions?
63411
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.
76888
Let $f: [0,\infty) \to [0,\infty)$ be a continuous function such that $f(f(x)) = x^2, \forall x \in [0,\infty)$. Prove that $\displaystyle{\int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}}$.  All I know about this function is that $f$ is bijective, it is strictly increasing*, $f(0) = 0, f(1) = 1, f(x^2) = (f(x))^2, \forall x \in [0, \infty)$ and $f(x) \leq x, \forall x \in [0, 1]$**. With all these, I am not able to show that  $\displaystyle{\int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}}$.  *Suppose that $f$ is strictly decreasing. Then, $\forall x \in (0,1), x^2 &lt; x \implies f(x^2) &gt; f(x) \iff (f(x))^2 &gt; f(x) \iff f(x) &gt; 1$, which is false because, if we substitute $x$ with $0$ and with $1$ in $f(x^2) = (f(x))^2$ we get that $f(0) \in \{0,1\}$ and $f(1) \in \{0,1\}$. So $f$ is strictly increasing.  **Suppose that there exists $x_0 \in [0, 1]$ such that $f(x_0) &gt; x_0$. Then, $x_0^2 = f(f(x_0)) &gt; f(x_0) &gt; x_0$, which is false. Then $f(x) \leq x, \forall x \in [0,1]$.  Edit:  I have come up with an idea to use Riemann sums, but I reach a point where I cannot continue.  Let $\epsilon &lt; 1$. Then $f(\epsilon) = x_1$ and $f(x_1) = \epsilon^2$. And now $(f(\epsilon))^2 = f(\epsilon^2) = x_2$ and so on. Now we will use the Riemann sum:  We will take the partition $\Delta = (1 &gt; \epsilon &gt; \epsilon ^2 &gt; ... \epsilon ^{2^n} &gt;0 ) $and the intermediate points will be the left margin of each interval. Then we have:  $\displaystyle{\int_{0}^{1}{(f(x))^2 dx}}  = \displaystyle{ \lim_{\epsilon \to 1}{\lim_{n \to \infty}{\sum_{k = 0}^{n}{(\epsilon^{2^k} - \epsilon^{2^{k+1}})\epsilon^{2^{k+1}}}}}}$  I do not know how to compute this.