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Let $f(t)$ be a convex function and define $g(t)$ to be the running average of $f(t)$

$$g(t) = \frac{1}{t} \displaystyle\int_0^t f(\tau) ~d\tau$$ Then $g$ is convex.

This is easy enough to prove just by differentiating twice. My question is: is there an easy geometric proof of this statement?

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A possibly useful change of variables: $g(t) = \int_0^1 f(ts)\,ds$. – Sean Eberhard Sep 11 '12 at 22:12
@SeanEberhard: Very nice. – copper.hat Sep 11 '12 at 22:15
up vote 2 down vote accepted

I would not differentiate twice, only because of the simple reason that you don't even know that $f$ is differentiable. However, there is a geometric way of looking at this. A convex function $f$ is defined by the following property: the graph of $f(x)$ between $x = a$ and $x = b$ lies below the line connecting the points $(a,f(a))$, $(b, f(b))$. I would like to say that pictorially I can think of the averages of the endpoints as always being larger than the full average over the whole interval $(a,b)$. So when you're averaging the average, this should remain the case. But maybe that isn't really precise enough...

Sean Eberhard's change of variables is very useful for formalizing this; for $r \in [0,1]$, $$ rg(a) + (1-r)g(b) = \int_0^1 rf(as) + (1 -r)f(bs) \, ds \geq \int_0^1 f(ras + (1-r)bs) \, ds = g(ra + (1 - r)b). $$

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