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Referring to this question on harmonic function in which is proved the sequent

Let $u(x)$ a harmonic function in $\mathbb{R}^n$ such as: \begin{equation} \int_{\mathbb{R}^n}|u(x)|dx =K< \infty \end{equation} Show thtat $u(x)=0$, $\forall x \in \mathbb{R}^n$.

Is there a heuristic argument to prove it or an explanation about why harmonic function that gives a finite value when integrated must be 0, without doing the rigorous proof?

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Yes there is. Heuristically, the "mean value over $\mathbb{R}^n$" of our function should give the value of our function at any particular point. Since the integral is finite and $\mathbb{R}^n$ is not, the function vanishes necessarily. This can be made rigorous as follows.

Supposing $u(p) = y \neq 0$ for some $p \in \mathbb{R}^n$, then the mean value theorem for harmonic functions and the triangle inequality give $$ |y| \cdot |B(p,R)| = \left|\int_{|x-p|\leq R} u(x) dx \right|< \int_{|x-p|\leq R} |u(x)| dx, $$ where $|B(p,R)|$ denotes the volume of the ball of radius $R$ around $p$. Thus, $$\int_{\mathbb{R^n}} |u(x)| dx = \lim_{R\rightarrow \infty} \int_{|x-p|\leq R} |u(x)| dx \geq \lim_{R\rightarrow \infty} y \cdot |B(p,R)| = \infty,$$ that is a contradiction.

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  • $\begingroup$ The argument "the "mean value over $\mathbb{R}^n$" of our function should give the value of our function at any particular point" can be used just because the function is harmonic right? Otherwise it should be valid for any integrable function which as a finite integral, and every function in $L$ would be 0... $\endgroup$ – opisthofulax Aug 16 at 9:30
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    $\begingroup$ @opisthofulax Oh yes, the mean value property is an important property of harmonic functions, and of course does not hold for a general function. $\endgroup$ – Josef E. Greilhuber 2 days ago

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