If the integrals of a harmonic function over horizontal lines are uniformly bounded, it is identically zero

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Let $u\colon\mathbb{R}^2\rightarrow\mathbb{R}$ be a harmonic function, such that $$\int\limits_{-\infty}^{+\infty} \lvert u(x,y)\rvert dx < C,$$ where $C>0$ is a constant not depending on $y\in\mathbb{R}$.

Is it true that such function must be identically equal to zero?

I tried to reduce this promblem to a case of Liouville's theorem, but it's not clear how to get uniform boundness on all the lines $\{ (x,y)\in\mathbb{R}^2 \colon\; y = \mathrm{const} \}$.

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Let's use the area form of the mean value property: For any $r>0,$

$$|u(0)| = |\frac{1}{\pi r^2}\int_{D(0,r)}u\,dx\,dy\,| \le \frac{1}{\pi r^2}\int_{D(0,r)}|u|\,dx\,dy$$ $$ \le \frac{1}{\pi r^2}\int_r^r\int_r^r|u(x,y)|\,dx\,dy\le \frac{1}{\pi r^2}\cdot C\cdot 2r = \frac{2C}{\pi r}.$$

Let $r\to \infty$ to see $u(0)=0.$ Clearly we could repeat the argument for a disc centered anywhere (or just consider the translates of $u$). Thus $u\equiv 0.$