(a) Show that there must exist $x,y \in [0,1] $ satisfying $|x-y| = \frac{1} {2}$ and $f(x) = f(y)$  I can start by defining a function $g(x) = f(x + \frac{1} {2}) - f(x)$ to guarantee an $x,y$ so that $|x-y| = \frac{1} {2}$ But how do I show that $f(x) = f(y)$?  (b) Show that for each $n \in \mathbb{N}$ $\exists x_n ,y_n \in [0,1]$ with $|x_n - y_n| = \frac{1} {n}$, and $f(x_n) = f(y_n)$  Actually I'm not sure where to start here. Any help is greatly appreciated. Thanks!
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Let $f \in C[0,1]$ and $f(0)=f(1)$.   How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$?  In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$.  For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous function $f \in C[0,1]$, such that $f(0) = f(1)$ and $f(a) \neq f(a+r)$ for any $a$.  This is called the Universal Chord Theorem and is due to Paul Levy.  Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question:     This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .
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Let $f$ a continous function defined in the interval $[0,1]$. Assuming: $\forall x \in [0,1]:f(x) &gt; x$ Prove: $\forall x \in [0,1]:f(x) &gt; x + \varepsilon $  I tried to use  and did some Algebra tricks but it didn't bring me to a safe shore :)    What do you suggest? Thanks!
97927
Prove that the function : $f(x) = x$ when $x$ is rational and $f(x) = 1 - x$ when $x$ is irrational is continuous at $x = \frac{1}{2}$  To prove the continuity it is necessary to prove that for every $\epsilon$ there exists a $\delta$ such that $|f(x) - f(\frac{1}{2})| &lt; \epsilon$ whenever $|x -c| &lt; \delta$ $- (1)$ Now if we choose $a  € (\frac{1}{2} - \delta, \frac{1}{2} + \delta)$ and also $a $ is rational and choose $\epsilon &lt; |a - \frac{1}{2}|$ then condition in $(1)$ is not satisfied. How to move from here?
118717
I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise.  Let $f: [0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$. Show that    for any $n\in \mathbb N$ there exists a horizontal closed interval of length $\frac 1n$ with endpoints on the graph of this function; if the number $\ell$ is not of the form $\frac 1n$ there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length $\ell$.   The first part can be proven like this: Consider $g: [0,(n-1)/n] \to \mathbb R$ given by $g(x) = f(x) - f(x+1/n)$. Then   $$\sum_{k=0}^{n-1} g(k/n) = 0$$  and therefore either all of these points are zero or there exists both a point where $g$ is positive and a point where $g$ is negative. By continuity, there must then also be a point where $g = 0$. So we are done.  Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for $\ell &lt; 1/2$.  (For $\ell &gt; 1/2$ the function $f(x) = \sin(2\pi x)$ will do.)  Can anyone help me out?  Cheers,
67634
Let $f:X\to Y$ be a function and $E\subseteq X$ and $F\subseteq X$.   Show that in general $f(E − F)\nsubseteq f(E) − f(F)$.   I have no idea about how to prove this; and could anyone please explain the basic theory of functions(by relating to this question. Thank you.
73022
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N} $ satisfying $\frac{1}{f(a)}+ \frac{1}{f(b)} = \frac{1}{f(c)} $ whenever $\frac{1}{a}+ \frac{1}{b} = \frac{1}{c} \ a,b,c \in \mathbb{N} $   After talking through it with a friend, they believe there is a simple solution only a couple of lines long, but I have yet to spot the method. Any ideas?
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Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.  I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x)+g(1)$. So if $g(1)=1$, we can show inductively that $g(n)=n$ for integer $n$. Maybe then extend to rationals and reals.
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?
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This is from Apostol's Mathematical Analysis.  Let $f$ be a function defined in $[0, 1]$ such that for each real number $y$ either there is no value of $x \in [0, 1]$ such that $f(x) = y$ or there are exactly two such values of $x$. Prove that a) $f$ must be discontinuous on $[0, 1]$. b) $f$ has an infinite number of discontinuities in $[0, 1]$.  Long back I had tried to solve the problem. I reasoned for a) as follows: Suppose that $f$ is continuous and let $m, M$ be its minimum and maximum values on $[0, 1]$. Clearly if $m = M$ then $f$ is constant and clearly for every $x \in [0, 1]$ we have $f(x) = m = M$ which contradicts given property of $f$. So we must have $m &lt; M$. Clearly there are now four distinct points $a, b, c, d$ in $[0, 1]$ such that $f(a) = f(b) = m, f(c) = f(d) = M$. By checking possible linear orderings of $a, b, c, d$ I could find a find a value of $y \in (m, M)$ for which there exist more than 2 values of $x$ such that $f(x) = y$. However this approach looks very clumsy and I suspect much simpler and easier proof is possible. Part b) seemed bit complex and my only guess was that probably $f$ was discontinuous on each subinterval of $[0, 1]$ but I could not prove this claim (may be the claim is wrong!). Perhaps there is a simpler solution to this part of the problem. Apostol's also asks to provide an example of such a function. Any hints on the overall problem will be appreciated.  Update: After looking at Marc's answer, I think I have found a way to show that there are infinitely many discontinuities. Earlier the solution was a part of the question itself but it made the question look quite lengthy hence I moved it to a separate answer.
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Here's the question:     Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function   such that $$\frac{1}{2y}\int_{x-y}^{x+y} f(t) dt=f(x)\text{ for all  }x\in\mathbb{R},y&gt;0$$ Show that there exist $a,b\in\mathbb{R}$ such   that $f(x)=ax+b$ for all $x\in\mathbb{R}$.     My attempt:  It immediately follows from Fundamental Theorem of Calculus that $f'(x)=\frac{1}{2y} \left( f(x+y)-f(x-y) \right)$ for all $x\in\mathbb{R},y&gt;0$.  Let $x\in\mathbb{R}$ and $y&gt;0$ be arbitrary. Then there is a $c\in(x-y, x+y)$ such that $\frac{f(x+y)-f(x-y)}{2y}=f'(c)$. But $\frac{f(x+y)-f(x-y)}{2y}=f'(x)$ hence,  $f'(x)=f'(c)$. But since $x$ was arbitrary, $f'(x)=f'(c)$ for all $x\in\mathbb{R}$. Hence, it follows again by Fundamental Theorem of Calculus that $f(x)=f'(c)x+f(0)$ for all $x\in\mathbb{R}$.    It appears to me as if my proof is correct but I never used the "twice differentiable part". Are there any flaws in my proof? Alternative proofs are welcome.    EDIT: As pointed out by , my proof is wrong. Hints would be appreciated.
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I need to use this theorem as part of a proof. Basically it goes as follows:  If $f(x_0) &gt; g(x_0)$ and $\frac{d}{dx} f(x) &gt; \frac{d}{dx} g(x)$ for all $x \geq x_0$, then $f(x) &gt; g(x)$ for all $x\geq x_0$.  What's the name of this theorem, if it has one? Or if it doesn't, is there a simple way to prove it that I could include in my proof? Thanks.
110715
Let $x_1, x_2, . .  , x_n$ real numbers from $[0, 1]$. Prove there is   $x \in [0, 1]$ so that $|x - x_1| + |x - x_2| + . . . + |x - x_n| =\frac n 2$     My attempt  Let $f:[0,1] \rightarrow R, f(x)=|x - x_1| + |x - x_2| + . . . + |x - x_n|$. Because $f$ continous, the image of $f$ is an interval. To prove $\exists x \in [0,1]$ so that $f(x)=\frac n 2$ it is enough to find two values $a,b \in [0,1]$ so that $f(a) \le \frac n 2 \le f(b)$. I failed to find these values.  Also, a solution without function continuity will be appreciated.
42594
I have a problem/proof that I can't quite solve.  Let $$f(x)=\begin{cases}x^2\quad &amp;x \in \mathbb{R} \setminus  \mathbb{Q}\\-x^2 \quad &amp;x \in \mathbb{Q}.\\ \end{cases}$$ Prove that $$\lim_{x\to n} f(x)$$ doesn't exist for $n$ other than $0$.  Maybe I can use the regular epsilon-delta proof? I tried that but wasn't sure of what assumptions I could make with the absolute values. I would appreciate any help.
85617
Theorem: Let $f:[0,1] \to [0,1]$. Assume $f$ is continuous. Then there exists   $c \in [0, 1]$ such that $f(c) = c$.         For any function $f$, let $g(x) = f(x) − x$. If the theorem is true, what does that say about $g(c)$?   Write a proof of the theorem. Do it by applying the Intermediate Value Theorem   in the appropriate way to the function $g(x) = f(x) − x$ with $a = 0$ and $b = 1$. Clearly state why the hypotheses of the IVT are satisfied here.        Well if the theorem is true, then there exists $c$ such that $f(c) = c$. Then $f(c) - c = 0$. Then $g(c) = 0$. Intermediate Value Theorem: Let $I$ be an interval and $f$ a function whose domain contains $I$. If $f$ is continuous, then for all $a, b \in I$ with $a &lt; b$ and all real numbers $k$, if $k$ is strictly between $f (a)$ and $f (b)$, then there exists $c$ such that $a &lt; c &lt; b$ and $f (c) = k$.   Proof Attempt: Take $a = 0$ and $b = 1$. Then $a,b \in[0,1]$ and $a &lt; b$. Since $f:[0,1] \to [0,1]$, it follows that for all real numbers $k$ such that $f(a) &lt; k &lt; f(b)$, then there exists $c$ such that $a &lt; c &lt; b$ so $0 &lt; c &lt; 1$ and $f(c) = k$. Then $f(c) - k = 0$. Thus, since $g(x)=f(x)-x$, it follows that $g(c) = 0$.   Looking to see if I did this correctly and/or if there is a more elegant way to prove problem 2. I am a little confused why the first theorem is true, but I went with it.
48826
Suppose that $f$ is continous on $[0,1]$ , differentiable on $(0,1)$ , and $f(0)=0$ and  $f(1)=1$.For every integer $n$ show  that there must exist $n$ distinct points $\alpha_1,\alpha_2\cdots,\alpha_n$ in that interval so that    $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $   Is it theorem or i can show it with basic differentiation knowledge .Please give me hint .
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This is from Putnam:     If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.   Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.  Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.  Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.  Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.
62941
Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ such that $$(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$$  I've tried subbing in heaps of values but I keep getting things like $f(0)=f(0)$ and other such useless results.  Any help would be hugely appreciated.