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The mean value theorem for integration says that, if $G$ is a continuous real-valued function defined over an interval, $G: [a,b] \to \mathbb{R}$, then the mean value of G on the interval is achieved as a certain point of the interval, i.e:

$$\exists x_0\in[a,b]: G(x_0) = \frac{1}{b-a} \int_a^b G(t) \, dt$$

Is this theorem true in two dimensions?

Let $G$ be a continuous, two-dimensional function defined over a connected, convex, closed subset of the plane, e.g the unit disc: $G: B^2 \to \mathbb{R^2}$. Is this true that the mean value of G on the disc is achieved as a certain point of the disc, i.e:

$$\exists (x_0,y_0)\in B^2: G(x_0,y_0) = \frac{1}{\text{Area}(B^2)} \int_{B^2} G(x,y) \, dxdy$$

? The Wikipedia page on Mean Value Theorem lists some generalizations, but I could not find this exact variant, which seems very intuitive.

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I originally read this question incorrectly and provided a proof (see below) for $G : D \to \mathbb{R}$ where $D$ is a connected, convex, closed subset of the plane.

There is a counter example for a mapping $G : D \to \mathbb{R}^2$. Let $K_1, K_2$ be two disjoint compact subsets of $D$ and suppose that the components of $G$, namely $G_1$ and $G_2$, are non-negative and compactly supported in $K_1, K_2$ respectively. Then a point satisfying the equality for $G_1$ must reside in $K_1$. The same holds for $G_2, K_2$. Since they are disjoint sets there cannot be a common point satisfying both equalities.

If $G$ was a mapping to the reals then it is true. Since $G$ is continuous on the closed set $D$, it must attain its minimum at a point $(x_0, y_0)$ and its maximum at a point $(x_1, y_1)$. Now it is true that $$G(x_0, y_0) \leq \frac{1}{\text{Area}(B^2)} \int_{B^2} G(x,y) \, dxdy \leq G(x_1, y_1)$$

Let $(x(t), y(t))$ be a line formed from $(x_0, y_0)$ to $(x_1, y_1)$. This line is in the set $D$ because $D$ is convex. Now, by the intermediate value theorem $G(x(t), y(t))$ attains all values from $G(x_0, y_0)$ to $G(x_1, y_1)$ so we are done.

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  • $\begingroup$ $G(x,y) \in \mathbb{R}^{2}$. $\endgroup$ – Marco Cantarini Jun 25 '15 at 11:43
  • $\begingroup$ @MarcoCantarini Thank you for pointing that out. $\endgroup$ – muaddib Jun 25 '15 at 12:14
  • $\begingroup$ Here it is probably good to mention that if the values lie in $\Bbb{R}^2$, then the theorem also fails if we only integrate over an interval. $\endgroup$ – PhoemueX Jun 25 '15 at 13:07
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In addition to muaddib's answer, I found the following specific counter-example. The domain is the square $[0,2\pi]\times[0,2\pi]$, and the function is:

$$G(x,y)=[sin(x+y), cos(x+y)]$$

Then, by symmetry it is easy to see that the integral of $G$ over the domain is $(0,0)$.

However, there is no point in which $G(x,y)=(0,0)$, because $|G(x,y)|$ is 1 everywhere.

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