Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be a $C^1$ function such that $\det(Df(x)) \neq 0$ and $|f(x) - x| &lt; \frac{1}{3}$ for any $x \in \overline{B}_1 (0)$. Prove that there is $p \in \overline{B}_1 (0)$ such that $f(p) = 0$.   I'm not sure about how to approach this problem (I tried to use sequences, since the ball is closed, the limit points are in the ball). I know that, probably, I must to use the Inverse Function Theorem. But I cannot see how the hypothesis $|f(x)-x|&lt;1/3$ is useful. I aprecciate any hints, but I don't want a complete solution.
120961
The question is two parts. Suppose $B$ denotes the closed unit ball in $\mathbb{R}^n$ and $U$ is an open set containing the ball. Let $f$ be a $C^1$, rank $n$ function $U \to \mathbb{R}^n$ such that $||f(x) - x|| &lt; 1/2$ for $x \in U$. First, show that $||f||^2$ attains a minimum in the interior of $B$. Second, show that $f$ actually hits $0$.  It is not so hard to see that the second claim implies the first. This is because $||f||^2$ cannot possibly be any smaller than $0$, so if $f(x) = 0$ for some $x$, this is as low as it goes. It is a minimum since $f$ is rank $n$, and thus a local homeomorphism by the inverse function theorem. This implies it is locally injective, and thus a local minimum. It must happen in the ball, since the inequality given says in fact it has to happen in the open ball of radius $1/2$.  It is a common theme among the writers of these questions that the first part be a special case of the second, which is often times provable without the first. Thus I'd like to find an argument which establishes this fact directly.  Any hints?
75614
Here is the question: Let $f:\mathbb{D}\rightarrow\mathbb{C}$ be an analytic function.  Suppose that there exists $\epsilon&gt;0$ such that for every $\theta$, $|\theta|&lt;\epsilon$, and $\lim_{z\rightarrow e^{i\theta}}f(z)=0$.  Prove that $f$ is identically $0$ on $\mathbb{D}$. My thoughts:  Well, honestly, I am quite stumped here.  I was thinking I could let $f$ be analytic in some other region (can I even do this?), say the lower half plane and suppose $\lim_{z\rightarrow x}f(z)=0$ whenever $-1\leq x\leq1$, by assumption.  Maybe then I could show that $f$ is identically $0$ on the bigger region, and then consider $f$ on the intersection of those two regions, and then ultimately making my way towards using the uniqueness theorem to show what I want... but I am quite stuck on how to do that... any thoughts, suggestions, etc. are greatly appreciated!  Thank you.
67021
Suppose $f\in L^{p}(\mathbb{R}^{n}) \cap L^{q}(\mathbb{R}^{n})$. How can I prove that for any $p \lt r \lt q$,  $$ \lVert f \rVert_{r} \leq (\lVert f \rVert_{p})^{(1/r-1/q)/(1/p-1/q)} (\lVert f \rVert_{q})^{(1/r-1/q)/(1/p-1/q)}\:? $$ I tried using Hölder to show that $f$ is in $L^{r}$ but I'm completely lost on how to go about it...
18219
I have a  basic question  Let $f$ be a $\mathcal{C}^1$  map  from  $[0,1]$ into $\mathbb{R}$ vanishing in $0$ and $1$. I want to prove that   $$\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 $$  and to see when we have equality.
106718
Let $f$ be a continuous function on the closed unit disk and holomorphic inside it, such that $f(z) = 0$ for all $ z = \exp (it) $ with $0 ≤ t ≤ \pi/2 $. Prove that $ f(z) = 0 $ for all $ z \in \overline{D}(0,1) $.  I have worked this exercise for absurdity, but it has not reached any conclusion. I would like to know a more sophisticated test, perhaps by a direct method.
111724
The Problem:   Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.  Prove that $f(x) = 0$ for all $x ≤ 0$.  (IMO $2011$ , Pr: $3$)  Here is the solution:  Before we prove that $f (x) = 0$ for all $x ≤ 0$, we will prove that $f (0) = 0.$ $\color{black}{\large\text{Case} \thinspace  1:}$ $f(0)&gt;0$ Let, $\varepsilon &gt;\dfrac{f(f(0))}{f(0)}$ and $x&lt;- \dfrac{\varepsilon+f(f(0))}{f(0)}$  we have, from $f(x)\leq xf(0)+f(f(0))$,  $f(x)&lt;-\varepsilon.$ Then from, $f(f(x))\leq f(x)f(0)+f(f(0))$ we have $f(f(x))&lt;0$. Finally,  since $0&lt;f(0)\leq f(f(0))$ ,we get $f(0)\leq-xf(x)+f(f(x))&lt;0$ which imply $f(0)&lt;0$, which gives a contradiction. $\color{black}{\large\text{Case} \thinspace  2:}$ $-\alpha \leq f(0)&lt;0$, where $\alpha\in\mathbb{R^+}.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have: $\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$. Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$. In this case, we have $f(x)&lt;0.$ Because, if $f(x)=0$ at least for some $x$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction. Then let, $x&lt;-\sqrt {\alpha}$,  from  $f(x + y) \leq yf(x) + f(f(x))$  we have: $$\begin{align} f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(f(x)) \leq (f(x)-x)f(x)+f(f(x))\Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x &lt;-\sqrt {\alpha} \Longrightarrow f(x)&lt; -\sqrt{\alpha}.\end{align}$$ Finally, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)&lt;-\alpha$, which gives a contradiction. So, we deduce that $f(0)=0$. Then, applying $f(0)=0$, from  $f(0)\leq-xf(x)+f(f(x))$ and $f(x) \leq xf(0)+ f(f(0))$ we have, $$\begin{cases} xf(x)-f(f(x)) \leq 0 \\ f(x) \leq 0 \\ f(f(x)) \leq 0 \end{cases} \Longrightarrow  \begin{cases} xf(x)\leq0 \\ f(x)\leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0, x\in\mathbb {R} \\ xf(x)\geq 0, x\leq 0 \end{cases} \Longrightarrow xf(x)=0, x\leq 0 \Longrightarrow \color{blue} {\boxed{ f(x)=0, \text{for all} \thinspace x\leq 0.}}$$ End of the Proof.  Can you verify this solution? Are there any mistakes, gaps, etc.? Thank you very much!
72460
Many tricky exercises concern the quest for functions that satisfy particular conditions. For example, let us consider the spaces $C_p( \mathbb R), 1 \leq p &lt; \infty$, of continuous functions on $\mathbb R$ such that $\int_{\mathbb R} \lvert f(x) \rvert^p \mbox{d}x &lt; \infty$. I found the following exercises rather demanding:   Find a function $f\in C_1(\mathbb R)$ such that $f$ is unbounded; Find a function $f \in C_1(\mathbb R) \backslash C_2(\mathbb R)$.   so one can deduce that $C_p \not\subset C_q$ if $p &lt; q$.  I'm quite sure that a function defined to be zero everywhere except for triangular peaks of height $k$ and base $1/k^3$ centered on positive integers $k$ is a solution for both the exercises. (it's a simple matter to give to this function an explicit form, but I think it would be rather unclear.)   Is it correct?  Can anyone give other examples? (for the one or the other exercise, not necessary a solution of both of them!)
123866
Take a continuous mapping $f: \bar{B^{n}} \rightarrow \bar{B^{n}}$, where $\bar{B^{n}}$ is a closed unit ball in $\mathbb{R}^{n}$. Assume that $f(x) \neq x$ for every $x \in \bar{B^{n}}$. Define another function $r$ by following the directed line segment from $f(x)$ through $x$ to its intersection with $\partial B^{n}$, and let the intersection point be $r(x)$. Is it immediately evident that $r$ is a continuous function? Is so how does it follow? Thanks.
37618
Is there an injective function $f :\mathbb{R}^2\rightarrow\mathbb{R}$?  I approached this problem from the perspective of $f^{-1}$, from which I showed there exists a surjective function $f^{-1}:\mathbb{R}\rightarrow\mathbb{R}^2$. Would this imply that there exists an injective function for the inverse mapping?
57591
Suppose $f: \overline{\mathbb{D}} \to \mathbb{C}$ is continuous, analytic in $\mathbb{D}$ and satisfies $|f(z)|&lt;1$ for $|z|=1$. Find the number of solutions to the equation $f(z) = z^n$ where $n$ is a positive integer. ($\mathbb{D}$ is the unit disk).  Some of my own attempts: Using the Maximum/Minimum Principle, we can split up the the problem in $2$ parts: Suppose $f$ is constant. It then holds that $z^n = c$ has exactly $n$ zeros because of the fundamental theorem of algebra.  Suppose $f$ is not constant. It holds that $f$ has at least one zero  in $\mathbb{D}$. It of course also holds that $z^n$ has a zero of multiplicity $n$ at the origin. I think I'm supposed to hit with stuff like Rouché's theorem, the Argument Principle or Schwartz lemma, but nothing seems to fit.   Thanks in advance.
67530
Let $\overline{B_d (x, r)}$ be the closure of an open ball, and let $\hat B_d (x, r)$ be a closed ball. Show that in any metric space $(X, d)$, $\overline{B_d (x, r)} \subseteq \hat B_d (x, r)$ for all $x \in X$ and $r &gt; 0$.    I used the definition of closure: $\overline{B}$ = $B \bigcup B'$. If $x \in \overline{B}$, then $x \in B$ or $x \in \overline{B}$, or both. If $x \in B$, then clearly $x \in \hat B$ since $B \subseteq \hat B$. But, if $x \in B'$, then $x$ is an accumulation point. Since $\hat B$ is closed, it contains all accumulation points of $B$. Thus, $x \in \hat B$.  Is this approach correct, or at least in the right direction? Should I be using the closure definition, or try another one? Thank you.
47716
$f:[-1,1]\to \mathbb{R}$ is a continuous function such that  $f(2x^2-1)=(x^3+x)f(x).$ Find $\lim_{x\to 0}\frac {f(\cos x)}{\sin x}$  My approach : Using the functional equation I got f is an odd function. $\implies f(0)=0$  let $g(x)=\frac{f(\cos x)}{\sin x},x\ne 0\implies |g(2x)|= \frac{1+\cos^2x}{2}|g(x)|\le |g(x)|$ $\implies |g(x)|\le |g\left(\frac {x}{2^n}\right)|...(1)$   Also By replacing $x$ by $\sin x$ in the given functional equation we get $g(2x)=-\frac{1+\sin^2x}{2}\frac{f(\sin x)}{cos x}...(2)$ From (1) and (2) we get, $|g(x)|\le \left|\frac {1+\sin^2 \frac{x}{2^{n+1}}}{2\cos \frac{x}{2^{n+1}}}f\left(\sin\frac {x}{2^{n+1}}\right)\right|$  lets $n\to\infty\implies |g(x)|\le 0\implies g(x)=0\,\,\forall x\in[-1,1]-\{0\}.$  $\implies \lim_{x\to 0}\frac {f(\cos x)}{\sin x}=0$
85287
Suppose $g\in C^1 [a,b]$. Prove that for all $\epsilon &gt; 0$, there is $\delta &gt; 0$ such that $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{&lt; \epsilon }$ for all points $c,d \in [a,b]$ with $0 &lt;|d-c|&lt; \delta$   First, I don't know what $C^1 [a,b]$ means.  Some ideas:  By Mean value theorem, $g'(c) =  {{g(b)-g(a)} \over {b-a}} $ since $c\in[a,b]$.   To show $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{&lt; \epsilon }$ whenever $0 &lt;|d-c|&lt; \delta$ for $ c,d \in [a,b]$. I guess I have to use the definition of limit of continuous function. But, I don't know how to connect all of these ideas.
88187
Let $$f\colon [0,1]\rightarrow [0, \infty )$$  be  continuous. Suppose that  $$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for  all }x\in[0,1].$$  Then   $A.$ No  such  function  exists.  $B.$ There  are  infinitely  many  such  functions.  $C.$ There is  only  one  such  function  $D.$ There  are  exactly  two  functions.  Now  I  was  thinking,  since $\int_{0}^{x} f(t) dt$  is  the  area  under  the  graph  of  $f(t)$  from $0$  to  $x$ so it  can  be  easily  made  greater  than  the  value  of  $f(t)$  at  one  point, namely  $x$.  So  there   will  be  infinitely  many  such  functions  satisfying  this  condition.   Is  my  reasoning  correct or  not?  Thanks.
94393
I have to prove that a function $f:]0,1] \rightarrow \Bbb R$ : $$ f(x) = \begin{cases} \frac1q,  &amp; \text{if $x \in \Bbb Q$  with $ x=\frac{p}q$ for $p,q \in \Bbb N$ coprime} \\ 0, &amp; \text{if $x \notin \Bbb Q $} \end{cases} $$  is discontinuous in every point $x \in \ ]0,1] \cap\Bbb Q$.  And then to consider $x \in \ ]0,1] \backslash \Bbb Q$ and prove that it is continuous.  For now I learned different ways to prove continuity (epsilon-delta, sequences), but I'm never sure what would be better to use in each different case.  I wanted to prove the discontinuity by using sequences: $$\forall x_n \quad x_n\rightarrow a \quad \Rightarrow \quad f(x_n) \rightarrow f(a)$$  I tried creating a sequence $ x_n=\frac1n + a$, we know it converges to $a$ but $f(x_n)\ $ doesn't  converges to $\ f(a)$ because there would still be some points not in our set (irrational numbers that creates gaps).  But I don't think it works, so I'm asking you if you could help me solving the two questions.
90957
I want to prove the following theorem (no idea whether it has a name):  Let $V = \mathbb{R}^n$ or $\mathbb{C}^n$ and $\|\cdot\|$ be a norm on $V$. Then, there exist $C_1, C_2 &gt; 0$ such that for all $x \in V$:  $$C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$$  I first let $x \neq 0$ (otherwise it would be trivial). Then, I divided by $\|x\|_\infty$ and normalized the vector $x$ such that $\|x\|_\infty = 1$. That left me with  $$C_1 \leq \|x\| \leq C_2$$  but I don't see how this could help me or how I could possibly limit an unknown norm. How can I proceed? Or is this the wrong way anyway? Thanks for any answers.
88689
I was told this question by a friend, who said that their friend had thought about it on and off for six months without any luck.  I have then had it for a while without any luck either.  It is in the style of a contest question, but it is unlike any other, as really nothing seems to gain any ground.  Here's the question:  Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function satisfying the following property:  If $ABC$ is the equilateral triangle with side lengths 1, we have for any point $P$ inside $ABC$, $f(\overline{AP})+f(\overline{BP})+f(\overline{CP})=0$ where $\overline{AP}$ is the distance from point P to vertex $A$.  (Example: by taking $P=A$ we see $f(0)=-2f(1)$.)  Prove or disprove: $f$ must be identically zero on its domain   Notice that continuity is critical since otherwise I could just take the function $f(x)=0$ whenever $x\in (0,1)$ and $f(0)=1$, $f(1)=\frac{-1}{2}$.