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Suppose $f \in C^2(\mathbb{R}^2)$ is harmonic. Suppose that $f$ is in one $L^p(\mathbb{R}^2)$ space, with $1 \le p \le \infty$.

How can I prove that $f$ is constant and in particular is identically $0$ if $1 \le p < \infty$ ?

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    Show that $f$ is bounded which you could use maximum principle. – 2017-01-14
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    Use the mean value property. – 2017-01-14
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    @Vik78 What do you have in mind? – 2017-01-14
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    By the way, in my answer below, I'm not interested in the bounty. @Glitch was first. I just wanted to simplify the approach. Jensen is more natural to me than Holder when an average is involved. – 2017-01-21

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I'm going to add to the proof by @Glitch because I think it can be simplified. Starting with the volume version of the mean value property (which follows from the spherical MVP), we have, for any $r>0,$

$$u(x) = \frac{1}{V(B(x,r))}\int_{B(x,r)} u\,dV,$$

where $V$ is Lebesgue volume measure on $\mathbb R^n.$ This implies

$$ |u(x)| \le \frac{1}{V(B(x,r))}\int_{B(x,r)} |u|\, dV.$$

For $1\le p <\infty,$ apply Jensen's inequality to see

$$\tag 1 |u(x)|^p \le \frac{1}{V(B(x,r))}\int_{B(x,r)} |u|^p\, dV.$$

Recall this holds for any $r>0.$ As $r\to \infty,$ $V(B(x,r)) \to \infty,$ while the integral in $(1)$ is bounded above by $ \int_{\mathbb R ^n} |u|^p\, dV < \infty.$ It follows that $|u(x)|^p = 0,$ i.e., $u(x) = 0.$ Since $x$ was an arbitrary point in $\mathbb R^n,$ we see $u\equiv 0$ as desired.

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    This is essentially the proof given by @Glitch, but Hölder's inequality was replaced by Jensen's inequality. – 2017-01-18
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    You did read my first sentence, didn't you? – 2017-01-18
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I'll present the proof with $1 \le p < \infty$ since the case $p=\infty$ can be handled with Liouville's theorem. I'll also work more generally in $\mathbb{R}^n$ for $n\ge 2$ rather than $n=2$, as you ask about.

The key is to use the mean-value property. It shows that for $x \in \mathbb{R}^n$ and $r>0$ $$ u(x) = \frac{1}{\omega_n r^{n}} \int_{ B(x,r)} u(y) dy $$ where $\omega_n = |B(0,1)|$. Then we use Holder's inequality:
$$ |u(x)| \le \frac{1}{\omega_n r^n} \int_{B(x,r)} |u(y)| dy \le \frac{1}{\omega_n r^n} (\omega_n r^n)^{1/p'} \Vert u \Vert_{L^p(B(x,r))} \\ = \frac{1}{\omega_n^{1/p} r^{n/p}} \Vert u \Vert_{L^p(B(x,r))} $$ where $1/p + 1/p' =1$. Since $u$ is such that $$ \Vert u \Vert_{L^p(\mathbb{R}^n)} < \infty, $$ we plug in to see that $$ |u(x)| \le \frac{1}{\omega_n^{1/p} r^{n/p}} \Vert u \Vert_{L^p(\mathbb{R}^n)}. $$ Sending $r \to \infty$ then shows that $$ |u(x)| \le \lim_{r \to \infty} \frac{1}{\omega_n^{1/p} r^{n/p}} \Vert u \Vert_{L^p(\mathbb{R}^n)}=0, $$ and so $u(x) =0$ for all $x$. Thus the only harmonic function in $L^p(\mathbb{R}^n)$ for $1 \le p < \infty$ is identically $0$.

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    But $C_{n,p}$ also depends on the measure of $B_n$ that goes to infinity as $r \to \infty$. – 2017-01-17
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    So I think the proof is faulty. – 2017-01-17
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    @Axel: No, also $1/r^n$ changes to $1/r^{np}$ and this is due to the measure of $B(x,r)$. – 2017-01-17
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    @gerw is correct, as is the proof. The constant $C_{n,p}$ depends only on $n$ and $p$. In fact, $C_{n,p} = \omega_n^{-1/p}$ where $\omega_n = \pi^{n/2} / \Gamma(1+n/2)$. – 2017-01-17
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    I've updated the proof so you can see the details of the form of $C_{n,p}$ more clearly. – 2017-01-17
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    You're right. Thank you for the explanation. One more thing: Liouville's theorem as I know it is about bounded functions, not about essentially bounded functions (that is, functions in $L^\infty$). How can we adapt it to our case? – 2017-01-17
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    If $u$ is harmonic then it's smooth, and so in particular it's continuous. A continuous function in $L^\infty$ is just a bounded and continuous function. In other words, bounded and essentially bounded are the same for harmonic functions. – 2017-01-18
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Since $f\in L^p$, we have that $f$ is a tempered distribution. Notice that $\Delta f =0$. Then $|\xi|^2\widehat f =0$, which implies that $\widehat f$ has support contained in {0}. Any distribution of support in $\{0\}$ is of the form $$\widehat f = \sum_{|\alpha|\leq d}c_\alpha \partial^{\alpha}\delta$$ where $d$ is a non negative integer and $c_\alpha \in \mathbb C$.

Taking the inverse Fourier transform we conclude that $f$ is an polynomial function.

But the only polynomial function in $L^p$ is $f=0$.