Let $f\in C(\mathbb{R})$  be a function that satisfies $\displaystyle\lim_{x\to\pm\infty}f(x)=+\infty$. Show that $f$ attains its absolute minimum, i.e. there is a point $x_0\in\mathbb{R}$ such that $f(x)\geq f(x_0)$ for all $x\in\mathbb{R}$.  I was thinking about studying the set $M=\{L\in\mathbb{R}: f(x)\geq L, \forall x\in\mathbb{R}\}$. But I'm not sure if this is the right way to proceed.
123563
If $f:\mathbb{R}\to\mathbb{R}$ is continuous and $$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\infty$$  Prove that $f$ has a minimum on $\mathbb{R}$; that is, there is an $x_m\in\mathbb{R}$ such that $$f(x_m)=\inf_{x\in\mathbb{R}}f(x)&lt;\infty$$  I think the best approach would be to prove the existence of a lower bound, that is $\exists M\in\mathbb{R}:M&lt;f(x),\forall x\in\mathbb{R}$, but I don't know how to do this. I am probably supposed to use the extreme value theorem or the intermediate value theorem, but I don't know how to do that.
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.
68947
Does there exist a  function $f:\mathbb R \to (0,\infty)$ such that $f(x)f(y)\le|x-y|, \forall x\in \mathbb Q , \forall y \in \mathbb R \setminus \mathbb Q$ ?
60902
The set of points where an arbitrary function $f: X \to \mathbb{R}$ defined in a set $X \subset \mathbb{R^m}$ admits a strict maximum or minimum is enumerable.  [Hint: Let $E \subset X$ be a dense enumerable subset. For each point $x \in X$ that is a strict maximum of $f$, choose $e_{x} \in E$ and $r_{x} \in \mathbb{Q}$ with $|e_{x} - x| &lt;r_{x}$ and $y \in B(e_{x}; r_{x}) - \{x\} \Rightarrow f(y) &lt; f(x)$. The application $x \mapsto (e_{x}, r_{x})$ is injective].  I have the following proof but I don't know if it is correct because $ f:X \to \mathbb{R}$ and $X \subset \mathbb {R^m}$. What do you think about this test?:  Let $M$ be the set of strict local minima of a function $f$. Then for every $p\in M$ there is an open interval $(a,b)$ such that $f(p)$ is smaller than $f(x)$ for any $x\in(a,b)\setminus\{p\}$. We can now choose rational numbers $a_p,b_p$ with $a&lt;a_p&lt;p&lt;b_p&lt;b$, and $f(p)$ will be the smallest value on $(a_p,b_p)$.  Define a map $g:M\rightarrow\Bbb Q^2$ taking $p$ to a pair $(a_p,b_p)\in\Bbb Q^2$ like above. It's clear $g$ is injection, since $f(x),f(y)$ cannot both be smaller than each other for $x\neq y$. Hence $g$ is an injection into a countable set showing $M$ is countable.  Could another demonstration be better than this one?
116901
Let  $f:X\times Y\to \mathbb{R}$. I could not come up with an example that satisfies $$\sup_{y\in Y}\inf_{x\in X}f(x,y)&lt;\inf_{x\in X}\sup_{y\in Y}f(x,y).$$ Any help would be appreciated. Thanks!
83190
Let $f$ be an entire function s.t $f(\mathbb{C}) \cap B_R(z_0) = \varnothing$ for some $z_0$ and some $R$. Then $f$ is constant.  I guess since the image of the whole plane isn't dense then $f$ doesn't have an essential singularity at infinity. Now I have to exclude it has a polar singularity at infinity..
56156
Question.  Suppose that a function $f$ is defined on an interval $I$, $c$ is a point of $I$, and {$x_k$} and {$y_k$} are any two sequences in $I$ such that $\lim_{k\to \infty}x_k=\lim_{k\to \infty}y_k=c$ with $x_k&lt;c&lt;y_k$ for all $k$. Prove that, if $f'(c)$ exists, then $\lim_{k\to \infty}\frac{f(x_k)-f(y_k)}{x_k-y_k}=f'(c).$   My Attempt.  Let $\epsilon_k=\frac{1}{k}$ be given.  For $\epsilon_1=1,$ there exists $\delta_1&gt;0$ such that $x_1,y_1\in I, x_1,y_1\in N'(c;\delta_1)$ implies $\vert \frac{f(x_1)-f(y_1)}{x_1-y_1} -f'(c) \vert&lt;\epsilon_1.$  For $\epsilon_2=\frac{1}{2},$ there exists $0&lt;\delta_2&lt;$min$\{\vert x_1-c \vert,\vert y_1-c \vert,\epsilon_2\}$ such that $x_2,y_2\in I, x_2,y_2\in N'(c;\delta_2)$ implies $\vert \frac{f(x_2)-f(y_2)}{x_2-y_2} -f'(c) \vert&lt;\epsilon_2.$  Similarly, for $\epsilon_{k}=\frac{1}{k},$ there exists $0&lt;\delta_k&lt;$min$\{\vert x_{k-1}-c \vert,\vert y_{k-1}-c \vert,\epsilon_k\}$ such that $x_k,y_k\in I, x_k,y_k\in N'(c;\delta_k)$ implies $\vert \frac{f(x_k)-f(y_k)}{x_k-y_k} -f'(c) \vert&lt;\epsilon_k.$  Then by taking the limit to infinity, we obtain $\lim_{k\to \infty}\frac{f(x_k)-f(y_k)}{x_k-y_k}=f'(c).$   It really took me a lot of time to finish the proof and I need this to be checked by an expert:) Is the proof okay? Any response would be appreciated!
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?
11660
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance.
117107
Let $X:=\{\chi \in H^1_{\text{loc}}(\mathbb{R})\}$ such that $\chi(0)=\frac {1}{2},\; \chi(-\infty)=(0)$ and $\chi(\infty)=1$.  Consider the functional $$F(\chi) = \int_{\mathbb{R}}(\chi'^2+\chi^2(1-\chi)^2)dx$$  Prove that $F$ has a minimizer on $X$ and compute it.  Hints: Let ($\chi_n)_{n\in \mathbb{N}}$ be a minimizing sequence.  We may assume $0 \le \chi_{n} \le 1$, why?  Can we assume that each $\chi_n$ is monotone?   Extract a subsequence that is weakly converging in $H^1_{\text{loc}}(\mathbb{R})$, i.e., weakly on each bounded interval of $\mathbb{R}$. Study lower semicontinuity.   Deduce the Euler-Lagrange equation in the weak form for the minimizer.  Does the candidate minimizer satisfy the conditions at $\pm\infty$?   Show that the minimizer $\chi$ is more regular. Integrate the equation to compute the explicit solution.  This method of solving such a problem is new to me so I don't really understand the hints, can somebody help me in any way?
97771
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. We denote, for every $x\in\mathbb{R}$, with $m(x)$ and $M(x)$ the minimum and maximum of $f$ on the interval $[x-1,x]$. Show that, if $m(x)+M(x)=0,\forall x\in\mathbb{R}$, then the functions $m$ and $M$ are constant.  I have tried to prove that $f$ is periodic, but I didn't managed to do it; I'm not sure it's even true...
119987
Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$?  If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.
83333
Hey guys I've been stuck on this problem for a while and I've still got no headway into how I'm supposed to do this. Any tips or guiding points would be helpful.  Let $f$ be a differentiable function such that $f′$ is continuous on $[0,1]$, and $M$ the maximum value of $|f′(x)|$ on $[0,1]$. Prove that if $f(0) = f(1) = 0$, then  $$\int_0^1 |f(x)| dx\leq\frac{M}{4}$$  I've come to a conclusion via Rolle's theorem that there exists a c between 0 and 1 where f'(c) = 0 and that would be a local extrema. But nothing past that other than the fact that the max value of M is 1.
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.
107447
Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.   $g(0)=1$. If $a=g(1)$,then $a&gt;0$ and $g(r)&gt;a^r$ for all $r$ in $\mathbb Q$.   Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.
6435
There is an exercise in my lists about those functions: $$f(x, y) = (y-3x^2)(y-x^2) = 3 x^4-4 x^2 y+y^2$$  $$g(t) = f(vt) = f(at, bt); a, b \in \mathbf{R}$$  It asks me to prove that $t = 0$ is a local minimum of $g$ for all $a, b \in \mathbf{R}$  I did it easily: $$g(t) = 3 a^4 t^4-4 a^2 t^2 b t+b^2 t^2$$ $$g&#39;(t) = 2 b^2 t-12 a^2 b t^2+12 a^4 t^3$$ $$g&#39;&#39;(t) = 2 b^2-24 a^2 b t+36 a^4 t^2$$  It is a critical point: $$g&#39;(0) = 0; \forall a, b$$  Its increasing for all a, b: $$g&#39;&#39;(0) = 2b^2 &gt; 0; \forall b \ne 0$$ and $$b = 0 \implies g(t) = 3 a^4 t^4$$ which has only one minimum, at $0$, and no maximum  However, it also asks me to prove that $(0, 0)$ is not a local minimum of $f$.  How can this be possible? I mean, if $(0, 0)$ is a minimum over every straight line that passes through it, then, in this point, $f$ should be increasing in all directions, no?
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.