$f: \mathbb{R}^n \rightarrow \mathbb{R},f$ is continious. $$f \text{ is convex} \Leftrightarrow f\left(\dfrac{x+y}{2}\right) \le \dfrac{f(x) + f(y)}{2}\ \ \   \forall x, y \in \mathbb{R}^n$$  One side comes straight from Jensen's inequality, while the other one is not obvious at all.   Maybe more divisions by 2 will help as I have heard.
16465
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 $(a,b)$.
102534
Suppose that $f:[a,b]\to\mathbb{R}$ be a twice-differentiable function, and that there exists $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Show that if $f''(x)&gt;0$ for all $x\in[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c&gt;\frac{a+b}{2}$.  Using Taylor's theorem, i solved the problem. But, i'd like to prove that using convexity. Give some comments or hints. Thank you!
91557
I'm trying to prove that  $$ \left| \frac{1}{x}-\frac{1}{y} \right| $$  is a metric on $X = \mathbb R^+$.  The properties of non-negativity and symmetry are obvious (do I actually even need to prove them?), but the tricky one is, as usual the inequality property, i.e.  $$ d(x,z) \le d(x,y) + d(y,z) \text{ for any } z \in X. $$  So, ok, I have that  $$ d(x,z) \le d(x,y) + d(y,z) \\ \iff \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| \\ \iff \left| \frac{1}{x} \right|- \left| \frac{1}{z} \right| \le \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| $$  ...and that's about as far as I've gotten–which is to say, not very far. I also tried multiplying both sides by $ \left| \frac{1}{x}-\frac{1}{z} \right| $, but that wasn't very fruitful. If someone could point me in the right direction, it'd be greatly appreciated. Thanks.
2383
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.
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.
109684
I want to prove it using Jensen inequality, so I need to prove that $g(x)=x\ln x$ is a convex function, which means  $$g\left(\frac{a+b}{2}\right)\le \frac{1}{2}\left(g(a)+g(b)\right).$$ How can I prove it?
72847
Let $G$ be a convex area and $f:G \rightarrow \mathbb C$ be a holomorphic function (with continious derivative). Prove: If $|f'(z)-1|\lt 1,$ then f is injective on $G$  Could anyone give me a hint ? I dont know how to tackle this problem
75875
Well If I've got a nice understanding, this proof involves the use of Rolle's and/or Mean Value Theorems. But I don't even an idea of how to start. We already know that  $$f\text{ concave}\Rightarrow f''&lt;0$$ but, then...
89932
For $a,b \in \mathbb R$, $p\geq2$ I try to show $$\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p.$$  Is this a popular inequality (At least I could not find it in the list of popular inequalities from wikipedia)? It seems to be related to convexity but I did not succeed to show it. A related inequality seems to be for $p \geq 1,a,b\geq0$  $$\left(\frac{a+b}{2}\right)^p\leq \frac{1}{2}a^p+\frac{1}{2}b^p,$$ which directly follows from the convexity of $x^p$ for positive numbers.
57059
I came across this problem :  Let $f\colon [a,b]\rightarrow \mathbb{R}^n$ a continuous vector valued function. Then it is true that:  $$\left\Vert\int \limits_a ^b f(t) dt\right\Vert \leq \int \limits_a^b \left\Vert f(t)\right\Vert dt$$  I consider that the integral of $f$ is given by integrating each component and then I am trying to reduce it to having an integral inequality for $n$ single valued functions but I am stuck.  I would appreciate some help.  Thank you!
5280
I have seen the following inequality  $$ \frac{x - y}{\log x - \log y} &gt; \sqrt{xy} \ , \quad \forall x&gt;y $$ be stated as a near "obvious" fact in another , on the site.  The inequality is very cute, but so far I have not been able to prove it.  It reminds me of the Lipschitz inequality, but has some minor differences. Also Jensens inequality comes to mind. Is there something obvious I am missing, or is this ineqality not as easy to prove as it looks?
57648
I need to prove the following statement     Let $S \subset \mathbb{R}^n$ a nonempty convex set and $f: S \to \mathbb{R}$. Then $f$ is convex in $S$ if and only if $f\left( \sum_{i=1}^m\lambda_ix_i \right) \leq \sum_{i=1}^m\lambda_if(x_i)$ for al $m \in \mathbb{N}$, for all $x_1, \dots, x_m \in S$ and for al $\lambda_1, \dots, \lambda_m &gt; 0$ such that $\sum_{i=1}^m\lambda_i=1$   My try: I think I managed to prove the backwads implication. Let $m=2$, and since $\lambda_1 + \lambda_2=1$ then $\lambda_2 = 1-\lambda_1$, so  $$f(\lambda_1x_1+\lambda_2x_2)=f(\lambda_1x_1+(1-\lambda_1)x_2) \leq \lambda_1f(x_1)+(1-\lambda_1)f(x_2)$$ since $f$ is convex in $S$ by assumption.  For the forward implication, I thought that induction might work. The case when $m=1$ is trivial. However, I'm struggling to prove the general case. Any help with that step will be highly appreciate. Thanks in advance!
108965
I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.
56401
I need a help with proving convexity  of $e^x$ function. $f(x)$ is a convex function, If it satisfies following condition.  $$f(c\cdot x_1+(1-c)\cdot x_2) ≤ c\cdot f(x_1)+(1-c)\cdot f(x_2), \quad    0 ≤ c ≤ 1$$
61624
Given a function on a closed interval $f\colon I\subset \mathbb{R}\to \mathbb{R}$ with $$f(x+y) \leq f(x) + f(y).$$ Moreover, I know that $f$ is    monotonic increasing continuous on all points except countably many continuous from the right such that the limit from the left at each point exists. $f(nx) \leq nf(x)$ for all $n\in\mathbb{N}$.   Can I say more on this function? In particular, I would like to show that $f(\lambda x) \leq \lambda f(x)$ for all $\lambda \geq 1$ or at least for all rational $\lambda \geq 1$.
64952
Could you please give me a hint to prove that $\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$.  I do not see how the critical case $n=2$ enters the picture.  Thank you!
48948
I'm trying to solve the following problem:  Let $f:K\rightarrow \mathbb{R} $, $f$ convex and $K \subseteq \mathbb{R}^n$ convex. Then $f$ is continuous on $K$.   I have proved the only case $n=1$, but for an arbitrary $n$??