- #1
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
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