$\require{mathtools}
%
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% Environments
\newcommand{\al}[1]{\begin{align}#1\end{align}} % need this for \tag{} to work
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% Greek
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\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
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% .. over and under
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% .. relations
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% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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% Complexity classes
% .. classical
\newcommand{\Poly}{\mathsf{P}}
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\newcommand{\PSPACE}{\mathsf{PSPACE}}
\renewcommand{\L}{\mathsf{L}}
% .. probabilistic
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% .. circuits
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Suppose we have some values $a_1, \dotsc, a_n$ with a fixed sum $\sum a_i$, and we care about some expression of the $a_i$’s. Then convexity can show that this expression is optimized when either:
- all the values are equal:
- for example, if $f$ is convex, then $\sum f(a_i) \ge n\,f\p{\f{\sum a_i}{n}}$;
- this is called Jensen’s inequality;
- all but one of the values are zero:
- for example, if $a_i \ge 0$ and $f$ is convex with $f(0)=0$, then $\sum f(a_i) \le f\p{\sum a_i}$;
- this is called superadditivity (or subadditivity when $f$ is concave and the inequality is in the opposite direction).
Since interesting functions are basically always either convex or concave (at least in the part that you’re interested in), these inequalities are applicable practically all the time.
How to use convexity
- Define the values $a_i$ in such a way that $\sum a_i$ is nice.
- Transform your expression to make a sum of the form $\sum f(a_i)$ appear inside.
- (for sub/superadditivity only) Ensure that $a_i \ge 0$ and $f(0)=0$.
- Plot $f$ to check its convexity, and get
- convex: $n\, f\p{\frac{\sum a_i}{n}} \stackrel{\text{Jensen's}}{\le} \sum f(a_i) \stackrel{\text{superadditivity}}\le f\p{\sum a_i}$;
- concave: $f\p{\sum a_i} \stackrel{\text{subadditivity}}\le \sum f(a_i) \stackrel{\text{Jensen's}}\le n\, f\p{\frac{\sum a_i}{n}}$.
If $f$ is strictly convex/concave, then equality happens iff all values are the same (for Jensen’s inequality), or all but one of the values are zero (for sub/superadditivity).
Jensen’s for expectations
When dealing with expectations instead of sums, Jensen’s inequality becomes slightly more elegant and general:
\[
\al{
\text{$f$ convex}\ &\implies \E[f(\BX)] \ge f\p{\E[\BX]}\\
\text{$f$ concave}\ &\implies \E[f(\BX)] \le f\p{\E[\BX]}.
}
\]
Subtracting the tangent
We can make Jensen’s inequality true termwise by subtracting the tangent of $f$ at the average point $\mu \ce \frac{\sum a_i}{n}$ from each term:
\[
\al{
\sum f(a_i)
&= \sum\p{f(a_i) - f'(\mu)(a_i-\mu)}.\\
%&\ge n \, f(\mu).
}
\]
#figure showing the space between the tangent and the curve
Indeed, this zeroes out the slope at $\mu$:
\[
\at{\frac{\d}{\d a_i}\p{f(a_i) - f'(\mu)(a_i-\mu)}}{a_i = \mu} = 0
\]
so if $f$ is convex, then $f(a_i) - f'(\mu)(a_i-\mu)$ is minimized when $a_i = \mu$, and therefore
\[
f(a_i) - f'(\mu)(a_i-\mu) \ge f(\mu).
\]
This is particularly useful when $f(\mu) = 0$ because sums of nonnegative values are delightful.
Special cases
- Jensen’s inequality
- sub/superadditivity