1. ## Mathematical induction

Hi. I need help with this:
Show that if f is concave up on [a,b] and if there exists numbers
$\displaystyle a\leq x_{j}\leq b$
$\displaystyle 1\leq j\leq p$
and
$\displaystyle 0\leq \lambda _{j}\leq 1$
$\displaystyle 1\leq j\leq p$
such that
$\displaystyle \sum \lambda _{j}=1$
then
$\displaystyle f(\sum ^{p}_{j=1}\lambda _{j}x_{j})\leq \sum ^{p}_{j=1}\lambda _{j}f(x_{j})$

True for j=2, prove for j=p+1

2. Originally Posted by Ralf
Hi. I need help with this:
Show that if f is concave up on [a,b] and if there exists numbers
$\displaystyle a\leq x_{j}\leq b$
$\displaystyle 1\leq j\leq p$
and
$\displaystyle 0\leq \lambda _{j}\leq 1$
$\displaystyle 1\leq j\leq p$
such that
$\displaystyle \sum \lambda _{j}=1$
then
$\displaystyle f(\sum ^{p}_{j=1}\lambda _{j}x_{j})\leq \sum ^{p}_{j=1}\lambda _{j}f(x_{j})$

True for j=2, prove for j=p+1

$\displaystyle j$ is not the induction variable $\displaystyle p$ is.

You will have proven a base case where $\displaystyle p=2$, now you need to show
that if this is true for $\displaystyle p=k$ for some $\displaystyle k$, then it is true for $\displaystyle p=k+1$.

RonL

3. I'm not sure how to show this. Can I maybe substitute
$\displaystyle \sum ^{k}_{j=1}\lambda _{j}x_{j}=\lambda _{u}x_{u}$
and then use the known case p=2 like this
$\displaystyle f(\sum ^{k+1}_{j=1}\lambda _{j}x_{j})=f(\sum ^{k}_{j=1}\lambda _{j}x_{j}+\lambda _{k+1}x_{k+1})=f(\lambda _{u}x_{u}+\lambda _{k+1}x_{k+1})\leq$$\displaystyle \lambda _{u}f(x_{u})+\lambda _{k+1}f(x_{k+1})=\sum ^{k+1}_{j=1}\lambda _{j}f(x_{j})$

I would really appreciate it if someone could give me a hint. If it's correct to do it the way I did it now (which i doubt), then how do I know if I can do the substitution I just did?