I have the following task:  Prove that if $ f : I \rightarrow \mathbb{R} $ is continuous ($ I $ is a range) and $$ \forall {x,y \in I} \qquad f\left(\frac{x+y}{2}\right) \leq \frac{f(x) + f(y)}{2} $$ then $f$ is a convex function.  Can somebody give a hint what can I use to prove this? This is a homework assignment, so I'd like to try solve it myself.
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Given that $$f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,$$ how can I show that $f$ is convex. Thanks.      Edit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.
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This is question B4 in 1999 Putnam Exam. I would appreciate if somebody could help me with this.      Let $f(x),f'(x),f''(x),f'''(x)&gt;0$ , $f'''(x)$ is a continuous function and $f'''(x)&lt;f(x)$ on $\mathbb{R}$. Then show that   $$f'(x)&lt;2f(x),~ \forall x\in \mathbb{R}.$$
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I have to prove that the function given below is continuous in $\mathbb R^2$:  $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$f(x,y)$= \begin{cases}  e^{-\text(\frac{1}{|x-y|})} &amp; \text{if $x\neq y$} \\ 0 &amp; \text{if $x=y$} \end{cases}   I have just started learning continuity in two/more dimensions ,so please help how to prove that the function is continuous on any path I take in 2 dimensions...
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Can you help me to prove that $$|x-y|\leq |x|+|y|$$ I get a proff of this equality, but it's very short and I don't know if it's correct.
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I was wondering if linear function is convex or concave? For example f(x)=x, is function whose second derivate is 0 so we cant tell anything using this criteria. Can someone help?
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Here is my problem.  Suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}\cup\{+\infty\}$ is a positive function. Define $f_n(x)=\inf \{f(y)+nd(x,y):y\in \mathbb{R}^n\}$. Prove that $f_n(x)$ is Lipschitz continuous and $f_n(x)\rightarrow f(x)$.  I find $f_n(x)$ are increasing, $f_n(x)\leq f(x)$ and for each $n$, $f_n(x)$ is upper semi-continuous  without using the condition $f\geq 0$. I wonder how this condition make sense. Thanks~
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Problem: Given $f:[0,1] \rightarrow \mathbb{R}$ ($f$ continuous ) and $f(x^2) = f(x)$ $\forall x \in [0,1]$. Show that function $f$ is a const.
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Suppose $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ and for all bounded measurable $f$, $$ \varphi\Big(\int_0^1fd\lambda\Big) \le \int_0^1\varphi(f)d\lambda $$ I'm asked to prove that $\varphi$ is a convex function. I have no idea how to even begin, only idea I've had is to try to suppose that $\varphi''(x)&lt;0$ for some $x\in(0,1)$ but then I haven't got a clue.
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Suppose that the function $f:[a,b]\rightarrow \mathbb{R}$ is continuous. For a natural number $k$, let $x_1,\cdots,x_k$ be points in $[a,b]$. Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$.    This can be proved by showing   $\hspace{130pt}f(a)&lt;\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}&lt;f(b)$  and using the intermediate value theorem to say there exists a point $z\in [a,b]$ such that the statement is true.  I'm not sure how to show this though. Any suggestions?
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Let $f:(a,b)\rightarrow \mathbb{R}$.   $f$ satisfied the following property:  If $\forall x_{1},x_{0},x_{2}\in(a,b)$ and $x_{1}&lt;x_{0}&lt;x_{2};$then$\frac{f(x_{0})-f(x_{1})}{x_{0}-x_{1}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}.$   My question : whether the function $f\in C((a,b))?$    We can get  $\displaystyle\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}. $Let $\displaystyle g(x)=\frac{f(x)-f(x_{0})}{x-x_{0}},$ if we can prove $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0});$ then we have $f^{'}_{-}(x_{0}) $ exists .In a similar way ,$f^{'}_{+}(x_{0}) $ exists.  $f^{'}_{-}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}-}f(x)=f(x_{0});$ $f^{'}_{+}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}+}f(x)=f(x_{0}).$  Obviously, $f$ is coutinuous at $x=x_{0}.$Further, $f\in C((a,b))!$     But  I failed to prove  $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0}).$Sometimes  I doubt the conculsion that  $f\in C((a,b)).$  Either make a counterexample to deny it ,or prove the conculsion is correct?
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Which functions $f:\mathbb{R} \to \mathbb{R}$ do satisfy $$f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$$ for all $x,y \in \mathbb{R}$  I think the only ones are of type $f(x) = c$ for some constant $c\in \mathbb{R}$ and the solutions of the Cauchy functional equation $f(x+y) = f(x)+f(y)$ and the sums and constant multiples of these functions. Are there other functions which are both midpoint convex and concave?
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Here's a homework question I'm struggling with:  Let $f,g$ two convex functions. Prove that $h(x)=\max\{f(x),g(x)\}$ is also convex  I don't know where to begin. The only thing I had in mind was was to try proving that if a function is convex on two sets $A$ and $B$, it is also convex on their union. That does not seem right though, for example, if I glue together $f(x)=x^2, g(x)=\frac{x^2}{1000}$ where $f$ is defined on $[0,1]$ and $g$ on $(1,2]$. Anyway, that was the only thing I thought about. Any better ideas? thanks!
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How can we show that a bounded and convex function on $\mathbb R$ is constant? Derivatives are of no use since the function does not have to differentiable. I saw an answer here I think a while ago but did not understand it at all.  Since derivatives are useless, we would have to use the definition and somehow show that the function lies between two values which are equal to each other. But I am unable to progress any further.
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My function is defined as: $$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} $$   I want to find out for which $a\in \mathbb{R}$ the function is continuous. I am not quite sure how to solve something like this. I know that for $a=1$ the function is definitely not continuous, which can be checked, since $(\frac1n,\frac1n)$ converges to $(0,0)$ but $f(\frac1n,\frac1n)$ does not. Now if I want $f$ to be steady in $(0,0)$, I figured that this is the case if $a&gt;1$. But this is where I get stuck. How to I prove or check, if for $a&gt;1$ the rest of the function is continuous as well? Any help is greatly appreciated!
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Suppose $f:\mathbb{R}^n\to \mathbb{R}$ is both convex and concave, how to prove that $f$ is linear? or exactly speaking, $f$ is affine?  I thought for the whole day, but I cannot figure it out.   When I was working on this problem, I met another problem, are all the convex function continuous? If not, is there any counter example?  Actually, I can prove for one dimensional case, in which $f:\mathbb{R}\to \mathbb{R}$. However, I cannot generalize it into n dimensional cases.  By the way, I use definition for convex(concave) like this: $$f(t\vec{x}+(1-t)\vec{y})\leq(or \geq) tf(\vec{x})+(1-t)f(\vec{y}), \forall t\in[0,1].$$  Thank you so much!
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I was wondering if this is true.     Let $f:(a,b) \rightarrow \mathbb{R}$ be a continuous real function and suppose $c \in (a,b)$ is the unique local maximum of $f$ and that $f$ has no other local extremum. Then $c$ is also an absolute maximum of $f$. The statement is analogous for minimums.    I have no idea how to prove this, nor have I thought of any counterexamples, though I am certain that the assumption of continuity is essential; otherwise consider something like $f:(0,\frac 32) \rightarrow \mathbb{R}$ given by  $$f(x) = \begin{cases} 1 - \frac x2 &amp;  x≤1 \\ \sqrt{\frac{1}{16} - (x-\frac 54)^2} &amp;  x&gt;1 \end{cases}$$  which admits a unique local maximum at $x=\frac 54$, no local minimum, but no absolute maximum.  Note that a similar question was asked  but the answers were not satisfactory and OP did not specify whether the interval was open or closed.
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Let $A \subset \mathbb{R}^n$ be closed and $d_A(x) = \inf_{a \in A} \|x-a\|$. I need to show that $d_A$ is convex if and only if $A$ is convex.  I have no good ideas in either direction. Can somebody please give me a hint?  For the direction ($\Longleftarrow$), my attempt was:  Let $a_1 := \operatorname*{argmin}_{a \in A} \|x-a\|, a_2 := \operatorname*{argmin}_{a\in A} \|y-a\|$, then:  \begin{align*} d_A(x) &amp;= \inf_{a \in A} \|\lambda x + (1 - \lambda)y - a\| = \inf \|\lambda(x-a) + (1-\lambda) (y-a)\|\\ &amp;\leq \lambda \|x-a_1\| + (1-\lambda)\|y-a_2\| = \lambda d_A(x) + (1-\lambda)d_A(y) \end{align*}  However, I'm not sure whether the inequality in the third step is actually correct.