# How do you show that an $L^p$ whole (holomorphic on the facility aircraft) function is $0$?

Simply to make clear, I intend to show that:

If $f$ is whole and also $\int_{\mathbb{C}} |f|^p dxdy <\infty$, after that $f=0$.

I assume I can show that this holds true for $p=2$, yet I'm not exactly sure concerning various other values of $p$ ...

0
2019-05-18 22:56:19
Source Share


If $f: \RR \to \RR$ is an analytic boundless Lebesgue integrable function, after that it has to specifically please this building (due to the fact that it is continual):. \ [ \ forall \ e, M > 0 \ ; \ exists x_0: |x | > |x_0 | \ Rightarrow \ lbg (\ > M \ ) < \ e,. \ ] where $\lbg$ is the common Lebesgue action.

After that you simply make use of that $f$ is analytic iff $\ forall \ message portable K \ part \ RR \ ; \ exists C > 0: x \ in K, n \ in \ NN \ Rightarrow \ left \ vert \ frac \ partial ^ n f \ partial x ^ n (x) \ appropriate \ vert \ leq C ^ n+1 n!$

But it is very easy to see that this is difficult given that $|f'|$ will certainly need to be randomly huge - or to put it simply that you can go much sufficient bent on the exactly on the x - axis that $|f|$ needs to go from being bigger than some large $M$ to smaller sized than some tiny $m$ on a really tiny period.

This is missing out on plenty of information, yet it needs to be rather very easy to fill in.

0
2019-05-21 11:41:28
Source

Right! Thanks, I really did not think of making use of Hölder's.

Basically,

$$|f(0)| = \left|\int_{|z|=R} \frac{f(z)dz}{z} \right| < \left(\int_{|z|=R} |f(z)|^p |dz|\right)^{1/p} \left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q}$$

And $\left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q} = R^{-1+1/q} (2\pi)^{1/q} = R^{-1/p} (2\pi)^{1/q}$

$$\int_0^\infty \left((2\pi)^{1/q} |f(0)| R^{1/p}\right)^p dR < || f ||_p^p <\infty$$

And the only means for this to fly is for $f(0)=0$, for $p\le 1$.

0
2019-05-21 10:19:05
Source

Use Hölder is inequality and also Cauchy is indispensable formula to show the function and also its by-products all disappears at absolutely no.

0
2019-05-21 09:40:16
Source