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Put a bit more precisely, suppose that $g:\mathbb C\to\mathbb C$ is entire and for some positive real $A,B$ satisfies ${\rm Re\,}g\,(z)\le A+B\,|z|$ for all $z\in\mathbb C$, does it follow existence of some $a,b\in\mathbb C$ with $g\,(z)=a+b\,z$ for all $z\in\mathbb C\,$? If above we instead had $|\,g\,(z)\,|\le A+B\,|z|\,$, then the conclusion would easily follow from the Cauchy formula for derivatives ${\rm D}^ig\,(0)$ for $2\le i\in\mathbb N\,$, but is there a simple argument for the above, or is it even true?

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The real part, $u(x,y)$, of $g(z)$ is a harmonic function, and these guys satisfy the mean-value property (i.e an integral formula for the function) which is somewhat reminiscent of Cauchy’s integral formula. Using this one can get bounds on derivatives of as well, similar to Cauchy’s estimates for holomorphic functions (which are obtained from the integral formula). In particular, if a harmonic function has at most linear growth, then it has to be a polynomial of degree $\leq 1$, i.e $u(x,y)=ax+by+c$.

Now, Cauchy-Riemann tells us the partials of the imaginary part $v(x,y)$, are constant, so $v$ is also at most degree $1$. Putting these together, $g(z)$ is of the form $\alpha z+\beta\overline{z}+\gamma$, but by holomorphy, $\beta=0$, so yes $g$ is affine.


Edit to elaborate:

Suppose $u:\Bbb{R}^n\to\Bbb{R}$ is harmonic and satisfies a bound of the type $|u(x)|\leq f(|x|)$, where $f:[0,\infty)\to [0,\infty)$ is a weakly-increasing function (like $A+Br$, which is the situation you have, with $n=2$). Then, applying the bound in the above link over the ball $B_R(0)$ tells us that for all multi-indices $\alpha$ with $|\alpha|=k$, \begin{align} |D^{\alpha}u(0)|&\leq\frac{C_k}{R^{n+k}}\|u\|_{L^1(B_R(0))}\\ &=\frac{C_k}{R^{n+k}}\int_{B_R(0)}|u(x)|\,dx\\ &\leq \frac{C_k}{R^{n+k}}\int_{B_R(0)}f(R)\,dx\\ &=\frac{C_k}{R^{n+k}}\cdot f(R)\cdot V_n\cdot R^n\\ &=C_k\cdot V_n\cdot\frac{f(R)}{R^k}, \end{align} where I have denoted the volume of the unit ball by $V_n$. So, if $\frac{f(R)}{R^k}\to 0$ as $R\to\infty$, then we see that all the $k^{th}$ partials of $u$ at the origin vanish. Thus, the only possible non-zero derivatives at the origin are those of order $\leq k-1$. Since $u$ is harmonic, it is real-analytic, and thus in a neighborhood of the origin, it must clearly be a polynomial of degree $\leq k-1$; but by uniqueness of analytic continuation, it is globally a polynomial of degree $\leq k-1$.

In your case, $f(R)=A+BR$, and so if $k\geq 2$, then $\frac{f(R)}{R^k}\to 0$ as $R\to\infty$, and thus we see that $u$ is a polynomial of degree $\leq 1$, exactly as claimed above.

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  • $\begingroup$ Thank you for the answer. You referred to a result that gives estimates for derivatives of (real) harmonic functions $u={\rm Re\,}f\,$. However, I didn't see there anything based only the premise $u\le\ldots$ I gave in my question. So your answer, as it now stands, does not actually answer my question. The link given above by $\endgroup$ Commented Apr 2 at 21:13
  • $\begingroup$ (cont) Riemann actually refers to the problem for which I need the answer and that I reduced to my question above. There the problem is reduced to the Borel–Carathéodory theorem whose proof in turn leans on the Schwarz lemma. So I now understand how the argument can be arranged. I thank you and "Riemann" for attention. I upvoted your answer without accepting it. $\endgroup$ Commented Apr 2 at 21:14
  • $\begingroup$ @The-unKnowN the $L^1$-norm in the link, which appears on the RHS of the estimate can certainly be bounded using your pointwise estimate. And in fact, the argument can be generalized a lot further; as long as you have some sort of bound $|u(z)|\leq f(|z|)$, then by imposing certain conditions on the asymptotics of $f(r)$ as $r\to\infty$, you get that such and such derivatives of $u$, say at the origin, vanish, and so with sufficient growth estimates on $f$, you can deduce $u$ must be a polynomial. $\endgroup$
    – peek-a-boo
    Commented Apr 2 at 21:32
  • $\begingroup$ If you only know $u(z)\le M$ for $z\in B$ how can you deduce even $\|u\|_{L^1(B)}<+\infty\,$? $\endgroup$ Commented Apr 2 at 21:43
  • $\begingroup$ @The-unKnowN see the edit. $\endgroup$
    – peek-a-boo
    Commented Apr 2 at 22:29

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