Families of holomorphic functions and uniform convergence on compact sets

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SUMMARY

The discussion focuses on the sequence of complex-valued functions defined as \( f_n = \tan(nz) \) for \( z \) in the upper half-plane \( \text{Im}(z) > 0 \). Two key properties are established: first, the sequence is uniformly locally bounded, with \( |f_n(z)| \leq M_0 \) for some \( M_0 \) and \( r_0 \) around any point \( z_0 \) in the upper half-plane. Second, the sequence converges uniformly to the constant function \( i \) on compact subsets of the upper half-plane. Corrections to the original analysis include ensuring the correct application of the decreasing nature of the \( \coth \) function and refining the convergence proof using the exponential form of the tangent function.

PREREQUISITES
  • Understanding of complex analysis, specifically holomorphic functions.
  • Familiarity with the properties of the tangent function in the complex plane.
  • Knowledge of uniform convergence and its implications on function sequences.
  • Proficiency in manipulating exponential functions and hyperbolic functions, particularly \( \coth \).
NEXT STEPS
  • Study the properties of holomorphic functions and their convergence criteria.
  • Learn about the behavior of the tangent function in the complex plane, particularly \( \tan(nz) \).
  • Explore uniform convergence in the context of sequences of functions and its applications.
  • Investigate the implications of the \( \coth \) function's monotonicity in complex analysis.
USEFUL FOR

Mathematicians, particularly those specializing in complex analysis, graduate students studying holomorphic functions, and researchers interested in uniform convergence properties of function sequences.

pantboio
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Consider the sequence $\{f_n\}$ of complex valued functions, where $f_n=tan(nz)$, $n=1,2,3\ldots$ and $z$ is in the upper half plane $Im(z)>0$. I want to show two facts about this sequence:
1) it's uniformly locally bounded: for every $z_0=x_0+iy_0$ in the upper half plane, ther exist $r_0,M_0>0$ such that $|f_n(z)|\leq M_0$ for every $n$ and for every $z$ with $|z-z_0|<r_0$.
2) the sequence converges uniformly to the function identically equal to $i$ on the compact subsets of the upper half plane.

For point 1), i write $tan(nz)$ in terms of exponentials:

$$tan(nz)=i\frac{e^{-inz}-e^{inz}}{e^{-inz}+e^{inz}}$$
Thus
$$|tan(nz)|=\frac{|e^{-inz}-e^{inz}|}{|e^{-inz}+e^{inz}|}\leq\frac{|e^{-inz}|+|e^{inz}|}{|e^{-inz}|-|e^{inz}|}=\frac{e^{ny}+e^{-ny}}{e^{ny}-e^{-ny}}=coth(ny)$$

Since $coth(y)$ is monotonically decreasing for $y=Im(z)>0$, we have
$$|tan(nz)|\leq coth(ny)\leq coth(y)\leq coth(y_0+r_0)=:M_0$$

Do you think there's some error in what i wrote?
For the point 2) i think i need a suggestion
 
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pantboio said:
Consider the sequence $\{f_n\}$ of complex valued functions, where $f_n=tan(nz)$, $n=1,2,3\ldots$ and $z$ is in the upper half plane $Im(z)>0$. I want to show two facts about this sequence:
1) it's uniformly locally bounded: for every $z_0=x_0+iy_0$ in the upper half plane, ther exist $r_0,M_0>0$ such that $|f_n(z)|\leq M_0$ for every $n$ and for every $z$ with $|z-z_0|<r_0$.
2) the sequence converges uniformly to the function identically equal to $i$ on the compact subsets of the upper half plane.

For point 1), i write $tan(nz)$ in terms of exponentials:

$$tan(nz)=i\frac{e^{-inz}-e^{inz}}{e^{-inz}+e^{inz}}$$
Thus
$$|tan(nz)|=\frac{|e^{-inz}-e^{inz}|}{|e^{-inz}+e^{inz}|}\leq\frac{|e^{-inz}|+|e^{inz}|}{|e^{-inz}|-|e^{inz}|}=\frac{e^{ny}+e^{-ny}}{e^{ny}-e^{-ny}}=coth(ny)$$

Since $coth(y)$ is monotonically decreasing for $y=Im(z)>0$, we have
$$|tan(nz)|\leq coth(ny)\leq coth(y)\leq coth(y_0+r_0)=:M_0$$

Do you think there's some error in what i wrote?
For the point 2) i think i need a suggestion
I think that the only error comes right at the end, where you have $\coth(y)\leqslant \coth(y_0+r_0)$. Since $\coth$ is a decreasing function, that should be $\coth(y)\leqslant \coth(y_0-r_0)$. If you then take $r_0 = y_0/2$, you see that you can take $M_0 = \coth(y_0/2)$.

For 2), start from $\tan(nz)=i\dfrac{e^{-inz}-e^{inz}}{e^{-inz}+e^{inz}}$, to show that $\tan(nz)-i =\dfrac{-2ie^{inz}}{e^{-inz}+e^{inz}} =\dfrac{-2i}{e^{-2inz}+1}$, and deduce that $\bigl|\tan(nz)-i\bigr| \leqslant \dfrac2{e^{2ny}-1}$. Then show that this goes to $0$ uniformly on compact subsets of the upper half plane.
 

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