<p>Holder's Inequality is pretty high powered for this problem :)</p>
<p>But, I wasn't really playing fair either.</p>
<hr>
<p>Definitions of convexity:</p>
<p>Def 1: A function $f$ is convex on an interval, if for all $a$ and $b$ in the interval, the line segment joining $(a,f(a))$ and $(b,f(b))$ lies above the graph of $f$.</p>
<p>Def 2: A function $f$ is convex on an interval if for $a$, $x$, and $b$ in the interval with $a<x<b$ we have
$\frac{f(x)-f(a)}{x-a}<\frac{f(b)-f(a)}{b-a}$</p>
<p>One of the problems on the assignment:</p>
<ol>
<li>Show that $f$ is convex on an interval if and only if for all $x$ and $y$ in the interval we have
$$f(tx+(1-t)y) < t\cdot f(x)+(1-t)f(y), ext{ for }0<t<1.$$
(This is just a restatement of the definition, but a useful one.)</li>
</ol>
<p>One of the study problems:
9)
Let $p<em>1,\ldots,p</em>n$ be positive numbers with $\sum^n<em>{i=1}p</em>i=1$</p>
<p>(c): Prove Jensen's Inequality: If $f$ is convex, then</p>
<p>$f\left(\sum^n<em>{i=1} p</em>i x<em>i\right)\le \sum^n</em>{i=1} p<em>i f(x</em>i).$</p>
<p>Hint: Use Problem 4, noting that $p_n=1-t$</p>
<hr>
<p>If you manipulate the original expression a little, you'll see that you have a straightforward application of Jensen. The trick is proving Jensen, which takes a little work (there's a problem for you).</p>
<p>My</a> proof of Jensen for my own indulgence</p>
<p>Way to blow by all that though, haha.</p>
<p>I didn't check my TeX code this time, so there may be a problem or two, but I must take notes for my 295 midterm tomorrow!</p>
