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How do I prove that $$\int \dot{x}^2 dt\geq \int \langle \dot{x}\rangle^2 dt $$

(i.e a free particle not in any external potential field movies with uniform veloctiy)

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Write out the full expression of $\langle \dot{x} \rangle$? In any case, is there anything other than the context which makes this a physics question rather than a math one? – dmckee Dec 3 '11 at 20:30

This is a standard convexity inequality. If a function has the property that its second derivative is always positive, it has the property that if you replace a,b with $a<b$ with the average $(a+b)/2$, then during the replacement f(a) goes up in value by less than the amount that f(b) goes down (just from the curving-up shape of the function), so that

$$ f(a)+f(b) \ge 2 f({a+b\over 2})$$

For any function f(x) with positive second derivative, like the squaring function. By repeatedly splitting a single value into the average of other values, you can prove the inequality for any collection of squared values, so that any average of squares is bigger than the average value squared, which reproduces the inequality you want in the continuous limit.

To prove the binary version of the inequality, let $m={a+b\over 2}$,

$$f(a) - f(m) = f'(x) (m-a)$$ $$f(b) - f(m) = f'(y) (b-m)$$

by the mean value theorem, where x is less than y, $m-a=b-m$ by the definition of the mean, and the inequality follows from convexity $f'(x)<f'(y)$.

This is the conceptual reason, an it is important for deriving other convexity inequalities, like Feynman's variational inequality for the exponential of the action, which uses the convexity of the exponential function in the same way as this argument used the convexity of the squaring function. But for this special case of a convex quadratic function, this is also a simple consequence of the positivity of the norm,

$$ \int |\dot{x} - A |^2 = \int |\dot{x}|^2 - 2 A \int \dot{x} + T A^2 > 0 $$

Which turns into what you want when A is equal to the mean value of $\dot{x}$,

$$A = {1\over T} \int \dot{x}$$.

Where T is the length of the integration domain, so that A is the total displacement over the total time.

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