Let $I \subseteq \mathbb R$ be an interval and $f:I \to \mathbb R$ increasing and convex. How can  I show that $f$ is bounded from below by an affine function, i.e. there exist $a \in \mathbb R$ and $b \in \mathbb R_+$ such that $a+bx \leq f(x)$ for all $x \in I$? What similiar statement would we have for concave function?
122399
A function $f : I \rightarrow \mathbb R$ on an interval $I$ is convex   if $f((1-t)x+ty)\le (1-t)f(x)+tf(y), \forall x,y \in I, t \in [0,1]$      Assume now that $I$ is an open interval.  Show that if $f$ is convex   then for each $c \in I$ there exists $m \in \mathbb R$ such that $m(x  − c) + f(c) \le f(x)$ for all $x \in I$ , and if in addition $f$ is   differentiable at $c$ then $f'(c)$ is the unique $m$ that works. In   general, must m be unique?   (we have previously shown that convex functions are continuous)  I think I'm missing the point of the question.  Surely, by the mean value theorem we can find some $y\in (x,c)$ such that $f'(y)=\frac{f(x)-f(c)}{x-c}$ setting $m=f'(y)$ we get $m=\frac{f(x)-f(c)}{x-c}, m(x-c)+f(c)=f(x)$ such an $m$ fits the inequality we're asked to show.  I also cannot see why $f'(c)$ would be the unique solution or why we would ever get a unique solution of the inequality at all for that matter.  Clearly I'm misunderstanding the question, so if someone could point out where I would appreciate that.  Thank you
60427
Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points then $\int^1_0f(x)dx$ exists.   I know that I can break up the interval $[a,b]$ into subintervals where each subinterval has one discontinuity point. Then if $f$ is integrable on each subinterval then it is integrable on $[a,b]$. So if I let $c$ be my point of discontinuity in the subinterval $[a_1,b_1]\subset [a,b]$, where $c\in[a_1,b_1]$. Then how can I show this subinterval is integrable?
91414
Show that $f(x)=||x||^p, p\ge 1$  is convex function on $\mathbb{R}^n$. I have tried to use Holder's inequality, but I still cannot solve this problem. Could you help me with this problem? Thank you so much.
79025
Trying to show that the set $S = \{(x_1,x_2):x_1x_2\geq4,x_1&gt;0,x_2&gt;0\}$ is convex. I tried proving by definition, let $x=(x_1,x_2)$ and $y=(y_1,y_2)$ and $\lambda\in[0,1]$ We want to show that $\lambda x + (1-\lambda)y \in S$ for $\lambda=1$ and $\lambda=0$ the answer is trivial. for $\lambda \in (0,1)$ I got a bit stuck What I tried was $\lambda x + (1-\lambda)y = \lambda(x_1,x_2) + (1 - \lambda)(y_1,y_2) = \lambda(x_1,x_2) - \lambda(y_1,y_2) + (y_1,y_2) = (\lambda x_1 - \lambda y_1 + y_1, \lambda x_2 - \lambda y_2 + y_2)$ So now if we can prove that $(\lambda x_1 - \lambda y_1 + y_1)\times(\lambda x_2 - \lambda y_2 + \lambda y_2) \geq 4$ we are done. I got stuck here, I tried looking for a way with the mean inequality but with no success. I think I am way off base, perhaps Jensen's inequality or something in that domain might be what I am looking for.
76185
How can I prove that this function is bounded using the given 2 hints? $$f(x) = \frac{x}{x^2 + 1},$$  Hints:   $(x + 1)^2 \geq 0.$ $(x - 1)^2 \geq 0.$   thanks!
57630
Let $f:\mathbb{R}\to\mathbb{R}$ be a function which is continuously differentiable.  Suppose that $\forall t\in\mathbb{R}$ it holds  $$f'(t)\leq 1-f(t)^2$$ Prove that $\forall t\in\mathbb{R}:|f(t)|\leq 1$.  I know about some comparison theorems, i.e. if we have a solution (in a compact domain) of the Riccati equation $\dot{x}(t)=1-x(t)^2$, with the same initial value as of $f$, then in this domain we get that $f\leq x$. My problem is that it only gives an upper bound and not a lower bound.
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I am trying to prove the following:  If $A\subset\mathbb R$ is bounded and $\,f:A\to \mathbb R\,$ is uniformly continuous, then $f[A]$ is bounded.  Could you check my proof?  Let $A \subseteq [-K,K]\subseteq \mathbb R$. Let $\varepsilon = 1$. If $f$ is uniformly continuous there is $\delta$ wuth $|x-y| &lt;\delta$ imply that $|f(x)-f(y)|&lt;1$ for all $x,y\in A$. Let $a \in A$. Then because $A$ is bounded there is a finite number of balls $B(a_n,\delta)$ that cover $A$. Let the number be $N$. Then $f(a)-N \le f(x) \le f(a) + N$ for all $x\in A$.
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Let $g: E^{m} \rightarrow E^{1}$ be a convex function, and let $h: E^{n} \rightarrow E^{m} $ be an affine function of the form $h(x)=Ax+b$, where $A$ is an $m \times n$ matrix and $b$ is an $m \times 1 $ vector. Then, show that the composite function $f : E^n \rightarrow E^{1} $ defined as $f(x)=g(h(x))$ is a convex function.      Also, assuming twice differentiability of $g$, derive the expression for the hessian of $f$
8515
Let $u:\mathbb R^2 \to \mathbb R$ be a non-surjective harmonic function.  $(i)$ Show that $u$ is bounded from below or from above.  $(ii)$ Prove that $u$ is constant (and therefore any harmonic function is constant or surjective)  I know how to prove $(i) \implies (ii)$:  Suppose $|u|\leq B$, by hypothesis, there exists an entire function $f$ such that $f(x+iy)=u(x,y)+iv(x,y)$. If I consider $e^{f(z)}$, then $e^{f(z)}$ is an entire function and $$|e^{f(z)}|=|e^{u+iv}|$$ $$=e^u|e^{iv}|$$ $$\leq e^B.1$$ $$=c \in \mathbb R_{\geq 0}$$  This proves $e^{f(z)}$ is bounded so, by Liouville's theorem, $e^{f(z)}$ is constant. In this exercise  it is proven that $e^{f(z)}$ constant $\implies$ $f(z)$ constant.  I am having problems with the first part ($(i)$) of the exercise. Suppose $u$ is not surjective, then there exists $u_0 \in \mathbb R$ such that $u(x,y) \neq u_0$ for all $(x,y) \in \mathbb R^2$. I have no idea how to deduce $u$ is not bounded from this hypothesis, I would appreciate any hints.
119882
I came across a problem that says:  Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function. If $|f'|$ is bounded, then which of the following option(s) is/are true?     (a) The function $f$ is bounded.   (b) The limit $\lim_{x\to\infty}f(x)$ exists.   (c) The function $f$ is uniformly continuous.   (d) The set $\{x \mid f(x)=0\}$ is compact.   I am stuck on this problem. Please help. Thanks in advance for your time.
82667
This is from Exercise 7 in p. 92 in Munkres's Topology.  Except for the trivial cases such as $Y$ is empty set or singleton, it seems if $Y$ is convex in an simply ordered set $X$ then $Y$ is interval or ray.  But I cannot start my proof because I cannot use $\sup$ or $\max$ functions.  What should I do?
108633
I am trying to show that if $f$ is convex in $(a,b)$ it is  in $[c,d]$ where $a \lt c \lt d \lt b$.  Here's what I have so far:  Let $t_1,t_2 \in \mathbb{R}$ such that $a \lt t_2 \lt c \lt d \lt t_1 \lt b$  and let $x_1,x_2 \in [c,d]$.  Because $f$ is convex I know that $$\dfrac{f(c)-f(t_2)}{c-t_2} \lt \dfrac{f(x_2)-f(x_1)}{x_2-x_1} \lt \dfrac{f(t_1)-f(d)}{t_1-d}$$  I think I'm almost there, but what is the right $C$?
121557
Let $f$ be an entire function, such that $$ \lvert\, f(x+yi)\rvert\le \frac{1}{\lvert y\rvert}, $$ for all $x$ and all $y\ne 0$.  Show that $f$ is bounded.
43481
PROBLEM. Let $f:[a,b]\to\mathbb R$ be a continuous function, such that $f'(x)\ge 0$, for all $x\in [a,b]\setminus A$, where $A\subset [a,b]$ is a countable set. Show that $f$ is increasing.  Attention. In this problem, we DO NOT assume that $f$ is differentiable in the whole $[a,b]$.  Notes. (1) If we assume that $f$ is differentiable in the whole interval, then we can easily show that $f'(x)\ge 0$, everywhere. For otherwise, if $f'(x_0)=c&lt;0$, for some $x_0\in [a,b]$, then by virtue of , $(c,0)\subset f'([a,b])$, and hence, $f'(x)&lt;0$, for uncountably many $x$'s.   (2) The conclusion of the problem does not hold if we replace the assumption $A$ is countable with $A$ is a set of measure zero. Take for example the , with a negative sign in front.  (3) If the hypothesis $f'(x)\ge 0$, is replaced by $f'(x)=0$, then the conclusion becomes f is constant.
117060
Let $V : \mathbb{R}^n \rightarrow  \mathbb{R}$ be (strictly) convex and continuously differentiable on $\mathbb{R}^n$. Show that for any $y \in \mathbb{R}^n$ and any $x \in \mathbb{R}^n$ $$ V(y) \geq V(x) + \nabla V(x)^\top(y-x) $$ For $n=1$ this implies that for any $x \in \mathbb{R}$ and for any $y \in \mathbb{R}$, $V(y)$ lies above the tangent line to $V$ at $x$.
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Let $f\colon\mathbb R \to \mathbb R$ be convex and $X_1,\ldots,X_n$ be i.i.d. random variables (e.g., uniform). Furthermore, suppose $g_1,\ldots,g_n\colon\mathbb R\to\mathbb R$ are increasing. Is it true that $$\mathbb E f(g_1(X_1) + g_2(X_1) + \cdots + g_n(X_1)) \ge \mathbb E f(g_1(X_1) + g_2(X_2) + \cdots + g_n(X_n))? $$ Like the previous version of this question (), which is the special case where the $g_i$ are all the identity, it seems like one approach is to show that $g_1(X_1) + \cdots + g_n(X_1)$ is a mean-preserving spread of $g_1(X_1) + \cdots + g_n(X_n)$. As before, I am also curious if there is a name for this sort of result/inequality.
3731
Given that a function $g(x)$ is a monotone increasing function. Its domain is an interval [c,d]. There are two sets $a_j \in A, j= 1,...,J$ and $b_k \in B,k=1,...,K$  on this interval, which satisfy $c_z \in C = B$\A and $c_z \geq a_j$ for all $a_j \in A$. For example, the interval $[c,d]=[0,8]$,$A=\{0,1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5,6,7\}$. In this case, does the following inequality hold? If yes, could you give me some hints to prove that?  $$\frac{1}{K} \sum_{k=1}^{K}g(b_k) \geq \frac{1}{J} \sum_{j=1}^{J}g(a_j)$$