A continuous function $f:S^1\to R$ then $\exists z\in S^1$ such that $f(z)=f(-z)$   My attempt:  On contrary suppose $f(z)-f(-z)\neq 0$ so either it is $&gt;0$ or $&lt;0$  WLOG $f(z)-f(-z)&gt;0$  But I could not proceed further .  Please give me hint.
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.
30995
If $f:[0,1]\to\mathbb R$ is continuous and $\int_0^x f(t)dt = ∫_x^1f(t)dt$ then prove that $f(x)=0$.  I suppose that $f(x)$ isn't zero, set $g(x)= \int_y^xf(t)dt$ with $y\in[0,1]$ and use Bolzano but I don't know if it's right.
100587
My function is defined as: $$f(x,y) =  \begin{cases}        \dfrac{xy^a}{x^2+y^2}&amp;(x,y) \neq (0,0)\\      \ 0&amp;(x,y) = (0,0)    \end{cases} $$   I want to find out for which $a\in \mathbb{R}$ the function is continuous. I am not quite sure how to solve something like this. I know that for $a=1$ the function is definitely not continuous, which can be checked, since $(\frac1n,\frac1n)$ converges to $(0,0)$ but $f(\frac1n,\frac1n)$ does not. Now if I want $f$ to be steady in $(0,0)$, I figured that this is the case if $a&gt;1$. But this is where I get stuck. How to I prove or check, if for $a&gt;1$ the rest of the function is continuous as well? Any help is greatly appreciated!
86744
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.
14432
Suppose that $f: \mathbb R \to\mathbb  R$ satisfies $f(x + y) = f(x) + f(y)$ for each real $x,y$. Prove $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb R$.  Proof: suppose $f$ is continuous at zero. Then let $R$ be an open interval containing zero. Then $f$ is continuous at zero if and only if $f(x) \to f(0)$ as $x \to 0$. Then $|f(x) - f(0)| &lt; \epsilon$. Can anyone please help me? I don't really know how to continue. Thank you.
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 $\: g(y) &lt; f(y) \:$?
60477
exercise: Suppose that $f:(0,∞) → R$  satisfies $f(x) - f(y) = f(x/y)$  for all $x,y$ in $(0, ∞)$ and $f(1) = 0$.   Prove f is continuous on $(0, ∞)$ if and only if f is continuous at $1$.    (proof→:) Suppose that $f:(0,∞) → R$ satisfies $f(x) - f(y) = f(x/y)$ for all $x,y$ in $(0, ∞)$ and $f(1) = 0$. Suppose f is continuous on (0,∞), then by definition f is said to be continuous at a point $a$ if for every $ε &gt; 0$  there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$.   Then suppose f is continuous at $1$, then by our defined function  f(x) - f(y) = f(x/y) , we get $|x-1| &lt; δ$ implies $|f(x) - f(1)| = |f(x) - 0| = |f(x)| =  &lt; ε$.   (proof ←) :Suppose that $f$ is continuous at $1$, then we need to show $f$ is continuous on every point $a$ on $(0, ∞)$. Thus, by definition, f is said to be continuous at a point $a$ if for every $ε &gt; 0$ there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$. Thus, $|x-a| &lt; δ → |f(x) - f(a)| = |f(x/a)| &lt; ε$. Then $f$ is continuous on on $(0, ∞)$.  Please can anyone please help me verify this makes sense. If not, can someone please give any feedback/hints. Thank you in advance.
73147
Let $f$ be a continuous real-valued function in a circle $|z|\leq1$ and $|f|\leq1$. Prove that $| \oint_{|z|=1} f(z)dz |\leq4 $.   Any hint, please.
52466
Let $f:[0,1]\to\mathbb{R}$ be continuous and $f(0)=f(1)$. Show that for any integer $n\geqslant 2$, there is $\xi\in(0,1)$ s.t.$f(\xi)=f\left(\xi+\frac{1}{n}\right).$  I think this requires the intermediate value theorem somewhere.
10490
Let $f:\mathbb R \to \mathbb R$ be a twice continuously differentiable function, with $f(0)=f(1)=f'(0) = 0$. Then   $f^{"}$ is the zero function. $f^{"}(0)$ is zero. $f^{"}(x)=0$ for some $x \in $ (0,1). $f^{"}$ never vanishes.   I don't know where to start, tried rolles theorem but cant catch the point..please help me to start..
81335
If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is an entire function such that $f(z)\neq w$ for all $z\in \mathbb{C}$ and for all $w\in [0,1]\subset \mathbb{R}$, how to prove that $f$ is constant (without using Picard's little theorem).   Any hint would be appreciated.
68668
If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $f(0)=f(1)$, then there exists $ x\in [0,1]$ such that  $f(x) = f(x+\frac{1}{n})$, where $n$ is any natural number.  Let $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$. Now, $f$ is continuous function on closed interval.So, it must attain its bounds. Let $m$ be minima at $c_1 \in [0,1]$ and $M$ be maxima at $c_2 \in [0,1]$.  Now if $c_1,c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $g(x)$ on $[c_1,c_2]$.  If $c_1$ or $c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $[c_1,c_1-\frac{1}{n}]$ or  $[c_2,c_2-\frac{1}{n}]$ respectively. If  $c_1,c_2$ both do not belong to $[0,1-\frac{1}{n}]$, then apply IVT to $g$ on $[c_1-\frac{1}{n},c_2-\frac{1}{n}]$.  Is this proof correct? Thanks.
8015
Be a continuous function $f:[0,1] \rightarrow [0,1]$.  Show that there is a $\zeta \in [0,1]$ with $f(\zeta)=\zeta$ ($\zeta$ is called fixed point).  Consider the function $g:[0,1] \rightarrow [-1,1]$, $g(x):= f(x)-x$.  $g$ is continuous.  Because of $f(0),f(1) \in [0,1]$ is $g(0)\geq0$ and $g(1)\leq 0$.   Because $f(0)$ has a value between $0$ and $1$, $f(0)\geq 0$.  $g(0) = f(0)-0= f(0) \geq 0 - 0 = 0$.  Because $f(1)$ has a value between $0$ and $1$, $ f(1) \leq 1 $.  $g(1)=f(1)-1 \leq 1 - 1 = 0$ $\Leftrightarrow g(0)\geq 0$ and $g(1)\leq 0 $  After the IVT: $\exists \zeta \in [0,1]:g(\zeta)=0 \Leftrightarrow f(\zeta) = \zeta $  $\zeta$ is a fixed point of $f$. $\Box$  My questions are:  Is this proof done in the correct way or have I missed something?  Is there something I can improve?
60820
Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?   Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?  Thank you.
42554
Find all the continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that if $x-y\in \mathbb{Q}$ then $f(x)-f(y)\in \mathbb{Q}$.   Can somebody give me some hint to solve this problem. I think that the functions are $f(x)=mx+b$ where $m\in\mathbb{Q}$.  Thanks!
80407
A student came to me with this question and I cracked my head for one hour but I couldn't unambiguously prove that it exists or it doesn't exist. Continuous and invertible function such that $f^{-1}(x) =1/f(x)$ on its domain of definition.
97943
Let $f(x)$ be a function such that $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$. Then for what value of $k$ is the relation $f(x + k) = f(x)$ necessarily true for every real $x$?  The answer/solution suggested in my module is like this: "this is a bit involved but can be proved that $k=6$". Could anybody explain me this?