Liouville's Theorem

From a Circle Formula to a Global Conclusion

Cauchy’s Integral Formula shows that values of a holomorphic function on a circle determine its values inside. Its derivative version gives an especially useful estimate: if the function is bounded on a circle, then its derivative at the center is bounded by a quantity inversely proportional to the circle’s radius. For a function holomorphic on the whole complex plane, circles centered at any chosen point can have arbitrarily large radii. If the function is bounded everywhere, this forces its derivative to vanish.

A function holomorphic on all of $\mathbb{C}$ is called entire. The result developed here, Liouville’s Theorem, is a striking rigidity statement about entire functions: global boundedness leaves no room for variation. We will use the Cauchy’s Integral Formula for the First Derivative from the previous tutorial, together with the Length Estimate for a Line Integral.

Definition (Entire and Bounded Entire Functions): A function $f:\mathbb{C}\to\mathbb{C}$ is entire if it is holomorphic at every point of $\mathbb{C}$. It is bounded if there is a finite constant $M\geq 0$ such that $|f(z)|\leq M$ for every $z\in\mathbb{C}$.

The word “entire” concerns where a function is holomorphic; “bounded” concerns the size of its values across its whole domain. A function may be bounded on each fixed disk without having one bound that works on the whole plane. That distinction will matter when we apply the theorem.

A Derivative Estimate on a Circle

Let $f$ be holomorphic on an open set containing the closed disk $\overline{D(a,R)}$. Applying the Cauchy’s Integral Formula for the First Derivative at the center $a$, and then the Length Estimate for a Line Integral, gives the following bound.

Theorem (Cauchy Derivative Estimate at the Center): Suppose $f$ is holomorphic on an open set containing $\overline{D(a,R)}$, where $R>0$, and suppose $|f(z)|\leq M$ for every $z\in C(a,R)$. Then $$ |f'(a)|\leq\frac{M}{R}. $$

Proof. The derivative formula gives

f'(a)=\frac{1}{2\pi i}\int_{C(a,R)}\frac{f(z)}{(z-a)^2}\,dz.

On the circle, $|z-a|=R$, so $|f(z)/(z-a)^2|\leq M/R^2$. The length of $C(a,R)$ is $2\pi R$. The Length Estimate for a Line Integral therefore yields

|f'(a)| \leq \frac{1}{2\pi}\left(\frac{M}{R^2}\right)(2\pi R) =\frac{M}{R}.

This proves the estimate. Notice that enlarging the circle improves the bound, provided the same upper bound $M$ is valid on that larger circle. $\square$

The location of the derivative and the center of the circle match in this version of the estimate. The circle must lie in a region where $f$ is holomorphic, and the bound $M$ must hold all along its circumference. For an entire function, the first condition is automatic for every finite radius.

Worked Example: Estimating a Derivative from Boundary Values

Suppose $f$ is holomorphic on a neighborhood of $\overline{D(2-i,5)}$ and satisfies $|f(z)|\leq 12$ on $C(2-i,5)$. The center is $a=2-i$, and the radius is $R=5$. The Cauchy derivative estimate applies directly:

|f'(2-i)|\leq\frac{12}{5}.

This conclusion uses only the bound on the circle, not a bound on the whole disk. It does not say that $|f'(2-i)|$ equals $12/5$; it says that the derivative’s modulus cannot exceed that value.

Liouville’s Theorem

Theorem (Liouville’s Theorem): Every bounded entire function is constant.

Proof. Let $f:\mathbb{C}\to\mathbb{C}$ be entire and bounded, so that $|f(z)|\leq M$ for every $z\in\mathbb{C}$, for some finite $M\geq0$. Fix any point $a\in\mathbb{C}$. For every $R>0$, the closed disk $\overline{D(a,R)}$ lies in $\mathbb{C}$, and $f$ is holomorphic on an open set containing it. The same global bound $M$ holds on $C(a,R)$. The Cauchy derivative estimate gives

|f'(a)|\leq\frac{M}{R} \qquad\text{for every }R>0.

If $M=0$, then $|f(z)|=0$ for every $z$, so $f$ is identically zero. If $M>0$, the right-hand side can be made arbitrarily small by choosing $R$ arbitrarily large. Since $|f'(a)|$ is nonnegative and is at most $M/R$ for every positive $R$, it follows that $|f'(a)|=0$. Thus $f'(a)=0$. The point $a$ was arbitrary, so $f'(z)=0$ for every $z\in\mathbb{C}$.

To conclude that $f$ is constant, take any $z_1,z_2\in\mathbb{C}$ and consider the line segment $\gamma(t)=z_1+t(z_2-z_1)$, $0\leq t\leq1$. The chain rule gives

\frac{d}{dt}f(\gamma(t)) =f'(\gamma(t))(z_2-z_1)=0.

The real and imaginary parts of $f(\gamma(t))$ therefore have zero derivative on this interval and are constant. Hence $f(z_1)=f(z_2)$. As the two points were arbitrary, $f$ is constant. $\square$

Worked Example: Determining a Bounded Entire Function

Suppose $f$ is entire and $|f(z)|\leq 4$ for every $z\in\mathbb{C}$, with $f(0)=2-i$. Liouville’s Theorem says that $f$ is constant. A constant function takes at every point the value it takes at $0$, so

f(z)=2-i\qquad\text{for every }z\in\mathbb{C}.

This conclusion is consistent with the stated bound, since $|2-i|=\sqrt{2^2+(-1)^2}=\sqrt{5}<4$. The value at one point determines the entire function once global boundedness and entire holomorphicity are known.

Consequences and Applications

Liouville’s Theorem can be applied indirectly: sometimes a function constructed from an entire function is bounded, even when the original function is not. For example, exponentiation turns an upper bound on the real part into an upper bound on the modulus.

Worked Example: An Entire Function with Bounded Real Part

Let $f$ be entire and suppose $\operatorname{Re} f(z)\leq 2$ for every $z\in\mathbb{C}$. Define $g(z)=e^{f(z)}$. The exponential and $f$ are entire, so their composition $g$ is entire. Moreover,

|g(z)|=|e^{f(z)}|=e^{\operatorname{Re} f(z)}\leq e^2.

Thus $g$ is bounded and entire, so Liouville’s Theorem makes $g$ constant. Differentiating gives $g'(z)=e^{f(z)}f'(z)$. Since $e^{f(z)}\neq0$ at every point and $g'(z)=0$, we obtain $f'(z)=0$ everywhere. The line-segment argument in the proof of Liouville’s Theorem then shows that $f$ is constant. The same reasoning works whenever the real part of an entire function has a finite upper bound.

Another useful strategy is to subtract two functions. If their difference is entire and bounded, Liouville’s Theorem forces that difference to be constant. This can determine the possible form of a function without requiring a direct formula for it.

Worked Example: A Bounded Difference Forces a Linear Form

Suppose $f$ is entire and there is a finite $M\geq0$ such that $|f(z)-z|\leq M$ for every $z\in\mathbb{C}$. Set $h(z)=f(z)-z$. Both $f$ and $z\mapsto z$ are entire, so their difference $h$ is entire. The assumed inequality says precisely that $h$ is bounded. By Liouville’s Theorem, $h$ is constant; write $h(z)=c$. Therefore

f(z)=z+c\qquad\text{for every }z\in\mathbb{C}.

Conversely, any function of this form has $f(z)-z=c$, whose modulus is bounded by $|c|$. Thus the condition does not merely give a necessary form: the functions $f(z)=z+c$ do satisfy it.

Worked Example: A Nonconstant Polynomial Is Not Bounded

Consider $p(z)=z^3-2z+4$. For any $z\in\mathbb{C}$, the reverse triangle inequality gives

|p(z)|=|z^3-2z+4| \geq |z|^3-|2z|-|4| =|z|^3-2|z|-4.

For positive real $r$, the lower bound $r^3-2r-4$ tends to infinity as $r\to\infty$. For instance, when $r\geq2$, $2r\leq r^2$ and $4\leq r^2$, so $r^3-2r-4\geq r^3-2r^2=r^2(r-2)$, which tends to infinity. Taking $z=r$ therefore shows that $|p(z)|$ is unbounded on $\mathbb{C}$. This does not contradict Liouville’s Theorem: the polynomial is entire, but it fails the theorem’s global boundedness hypothesis.

Why Global Boundedness Matters

Liouville’s Theorem is not a claim that every holomorphic function is constant. For example, $f(z)=z$ is entire and nonconstant, but $|f(z)|=|z|$ grows without bound. The theorem’s two hypotheses work together: holomorphicity everywhere allows the derivative estimate on circles of any radius, while one global bound supplies the same $M$ on all those circles. If the available bound increases with the radius, the estimate may not force the derivative to vanish.

This is also why being bounded on each disk separately is insufficient. Every continuous function is bounded on a fixed closed disk, but the bound may depend on the disk. Liouville’s proof requires a single finite $M$ that works for every $z\in\mathbb{C}$. The inverse-radius estimate then makes $|f'(a)|$ no larger than $M/R$ for arbitrarily large $R$.

The theorem is a bridge from local complex analysis to global conclusions. Its proof uses the local-looking derivative formula, but the freedom to enlarge the circle turns that formula into a statement about the whole plane. The next tutorial uses Liouville’s Theorem in a major existence argument for polynomials.

Check Your Understanding

Use the derivative estimate and Liouville’s Theorem to answer the following questions.

Why does an entire function permit the derivative estimate to be applied on circles of every positive radius centered at a fixed point?
In the proof of Liouville’s Theorem, where is the assumption of one global bound used?
If $|f(z)|\leq 6$ on $C(a,3)$, what upper bound does the Cauchy derivative estimate give for $|f'(a)|$?
How can an upper bound on the real part of an entire function be converted into a bounded entire function?
Why does $p(z)=z^3-2z+4$ not contradict Liouville’s Theorem?

What You'll Learn