Recent content by Jose27

Just a thought (I don't know if it'll work), but maybe using the $\Gamma$ function's integral representation (remember we have to split the integral several times in order to extend it past a certain integer) and using Fubini's theorem. Something along these lines.

I think there's something wrong: Take $z=\frac{1}{2}$ (the estimate works for any positive real number on the unit interval) then for $n>2$ we have $|(1+nz)|^n \geq 1+(nz)^n>2$ which cannot happen if the product converges since individual terms tend to $1$ in that case. As for your other...

The product converges normally: Pick $K=D(0,r)\subset D(0,1)$ and let $|\cdot|_K=\sup_K |\cdot|$ then $$\sum_{n=1}^{\infty} |nz^n|_K\leq r|(\frac{1}{1-r}-1)'|=\frac{r}{(1-r)^2}<\infty$$ for each $0<r<1$. This implies that $f$ is holomorphic in $D(0,1)$

For a use the mean value property for harmonic functions. For b use the maximum and minimum principle.

Look up the identity theorem. For the rest, surely you can argue that if $f^{(m)}\equiv 0$ then $f$ is a polynomial.

I think this works, it's basically the same approach as with classical ODE's IVP: Define the operator $A:L^\infty[t_0,t_1] \to L^\infty[t_0,t_1]$, with $t_1>t_0$ to be defined, as $A(x)(t)=x_0+\int_{t_0}^{t_1} p(s)x(s)+f(s)ds$. It's easy to see $A$ is well defined as a mapping between these...

Using the method of characteristics you shoudl arrive at something like \(u(x,t)=\phi(y_0(x,t))\) where \(y_0(x,t)\) is the intersection of the \(x\)-axis with the characteristic passing through \((x,t)\). For this method I would recommend John's book "Partial differential equations" where he...

As long as you know why each of your claims is valid then yes, everything's fine.

One question: Do you know about the Phragmén-Lindelöf principle (look it up in Wikipedia, if you're not familiar with the name)? If so, this is just a special case since \(e^{\sqrt{x}}\leq e^{\sqrt{|z|}}\).

This is false as written: Take \(f(z)=a(z-1)\) for \(a\in \mathbb{R}\) small enough. This is because your inequality implies that \(f(0)=0\) which need not happen.

I won't write out everything, since apparently my definition of the Fourier transform is different (non-essentially though) from yours. You should arrive at an equation of the form \(\hat{u}_t (x,t)= c(x)\hat{u}\) (\(c(x)\) will depend on your definition of the transform) with the initial...

Since you have a representation formula for \(u\), everything's easier. You have $$u(x,t)= \frac{1}{\sqrt{4\pi t} } \int_{\mathbb{R}} e^{\frac{(x-y)^2}{4t}}g(y)dy$$ This gives, using Hölder's inequality with \(p=1, p'=\infty\) $$|u(x,t)|\leq \frac{1}{\sqrt{4\pi t} } \int_{\mathbb{R}}...

Take \(D'\subset D\) any closed subdisk, and consider the sets \(A_n=\{ x\in D' : f^{(n)}(x)=0 \}\). Prove that one of these, say \(A_k\), has an accumulation point in \(D\), what can you say about \(f^{(k)}\)?

This is not true: Take \(f(z)=e^{z^2}\) then \(f=1 on \partial S\) but it's unbounded. This is an application of the Phragmén-Lindelöf theorem, it shouldn't be too hard to find a good text with a proof.

I think you're dealing with something similar to this: \frac{\partial f}{\partial \overline{z} } = \frac{1}{2} \left( \frac{\partial f}{\partial x } +i \frac{\partial f}{\partial y } \right) and so a function f is analytic iff \frac{\partial f}{\partial \overline{z} }=0 . This is sometimes...