Let z(x,y) be continuous in a neighborhood of the origin.
Let $z(x,y)$ be continuous in a neighborhood of the origin. Prove that: $$ \lim_{\varrho\to0}\frac{1}{\pi\varrho^2}\iint_{x^2+y^2\leqslant\varrho^2} z(x,y)\,dx\,dy = z(0,0). $$
Explanation of Method
The quantity $\dfrac1{\pi\varrho^2}\displaystyle\iint_{x^2+y^2\leqslant\varrho^2}z\,dx\,dy$ is the average value of $z$ over the disk of radius $\varrho$ centered at the origin (since $\pi\varrho^2$ is the area of that disk). Continuity of $z$ at the origin means $z$ is nearly constant, equal to $z(0,0)$, on small disks — so the average over a shrinking disk should converge to that central value. This is made precise with an $\varepsilon$-$\delta$ argument bounding the deviation of the average from $z(0,0)$.
- Subtract the constant $z(0,0)$ inside the integral, using that it integrates to $\pi\varrho^2\cdot z(0,0)$.
- Bound the remaining integral using continuity of $z$ at the origin.
- Conclude via the definition of the limit.
Step 1 · Isolate the difference
Let $D_\varrho=\{(x,y):x^2+y^2\leqslant\varrho^2\}$, with area $\left|D_\varrho\right|=\pi\varrho^2$. Since $\displaystyle\iint_{D_\varrho}z(0,0)\,dx\,dy=z(0,0)\cdot\pi\varrho^2$, we may write:
$$ \frac{1}{\pi\varrho^2}\iint_{D_\varrho}z(x,y)\,dx\,dy-z(0,0) = \frac{1}{\pi\varrho^2}\iint_{D_\varrho}\big[z(x,y)-z(0,0)\big]\,dx\,dy. $$
Step 2 · Use continuity at the origin
Let $\varepsilon>0$. Since $z$ is continuous at $(0,0)$, there exists $\delta>0$ such that:
$$ x^2+y^2<\delta^2 \implies \big|z(x,y)-z(0,0)\big|<\varepsilon. $$
So for every $\varrho<\delta$, the bound $\big|z(x,y)-z(0,0)\big|<\varepsilon$ holds throughout $D_\varrho$.
Step 3 · Bound the averaged difference
For $\varrho<\delta$:
$$ \left|\frac{1}{\pi\varrho^2}\iint_{D_\varrho}\big[z(x,y)-z(0,0)\big]\,dx\,dy\right| \leqslant \frac{1}{\pi\varrho^2}\iint_{D_\varrho}\big|z(x,y)-z(0,0)\big|\,dx\,dy < \frac{1}{\pi\varrho^2}\cdot\varepsilon\cdot\pi\varrho^2 = \varepsilon. $$
Step 4 · Conclude
We have shown: for every $\varepsilon>0$ there is a $\delta>0$ such that $\varrho<\delta$ implies
$$ \left|\frac{1}{\pi\varrho^2}\iint_{D_\varrho}z(x,y)\,dx\,dy-z(0,0)\right|<\varepsilon. $$
This is precisely the statement that:
$$ \lim_{\varrho\to0}\frac{1}{\pi\varrho^2}\iint_{x^2+y^2\leqslant\varrho^2}z(x,y)\,dx\,dy = z(0,0). $$
Additional Notes
- An equivalent, more geometric argument uses the Mean Value Theorem for double integrals: since $z$ is continuous on $D_\varrho$, there is a point $(\xi,\eta)\in D_\varrho$ with $\displaystyle\iint_{D_\varrho}z\,dx\,dy=z(\xi,\eta)\cdot\pi\varrho^2$, so the average equals $z(\xi,\eta)$ exactly. As $\varrho\to0$, $(\xi,\eta)\to(0,0)$ (it lies within distance $\varrho$ of the origin), and continuity gives $z(\xi,\eta)\to z(0,0)$.
- This result is the two-dimensional analogue of the fact that a continuous function equals the limit of its averages over shrinking neighborhoods — the double-integral version of the Lebesgue differentiation theorem's continuous case, and the key identity underlying mean-value characterizations of harmonic functions in the plane.
Result
This establishes the claim: