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Mar
6
answered If $f$ is a holomorphic function in a rectangle in the first quadrant, and $|f(z)| \leq Re(z)$, prove that $f = 0$ for all the rectangle.
Mar
2
comment Show that complex function with a maximum at two values has a maximum at an intermediate value
@SrinivasK: Actually, in some cases $g(z)$ won't be holomorphic at $0$, because it's possible that $f(0)\neq 0$. But nonetheless, $g(z)$ is holomorphic on $\mathbb{C}^*$. In particular, it is holomorphic on the annulus $1\leq |z|\leq 10$. And on both boundary circles of the annulus, we are given that $|g(z)|< 1$. Since this is true on the boundary of the annulus, the maximum principle should tell you it's true everywhere inside the annulus as well. In particular, $|g(z)|<1$ whenever $|z| = 5$, for instance.
Mar
2
comment Show that complex function with a maximum at two values has a maximum at an intermediate value
Also, for $f(0)$, you know by the maximum principle that $|f(0)|<1$, but a good question to ask is if $|f(0)|$ can be as close to $1$ as you like. For this, I suggest even just looking at linear polynomials $f(z) = a + bz$. Can you find examples of such functions where $|a|$ is very close to $1$ and $f$ satisfies the desired estimates?
Mar
2
comment Show that complex function with a maximum at two values has a maximum at an intermediate value
Hint: Consider the function $f(z)/z$. It's holomorphic away from the origin, and you have estimates on its size along the circles $|z|=1$ and $|z|=10$, so you should be able to conclude something about all the circles $|z|=r$ for $1<r<10$.
Mar
1
answered Compact normal family
Jan
26
awarded  Yearling
Jan
3
awarded  Nice Answer
Jan
1
comment Prove linearly independence of column vectors
If $Au_0$ is $0$, what does that say about $A^{k-1}u_0 = A^{k-2}(Au_0)$?
Sep
30
awarded  Explainer
Sep
11
awarded  Enlightened
Sep
11
awarded  Nice Answer
Jul
14
answered Proving Cauchy inequality involving four expression
Jul
14
comment Proving Cauchy inequality involving four expression
As stated, the inequality you write is not true. For instance, it is false when you take $a = b = c = 1/3$. Perhaps there's a typo, or an extra hypothesis. Do you mean $(a^2 + b^2 + c^2)$ in place of $(a^4 + b^4 + c^4)$?
Jul
13
answered If $f'(x)\cdot x$ goes to zero then $f(2x)-f(x)$ is bounded.
Jul
6
comment Irrational Rotation
@Priyanka: density is a topological property, and all topological properties are preserved under homeomorphism.
Jul
4
answered Irrational Rotation
Jul
2
awarded  Curious
May
6
comment Topology of the tangent bundle of a smooth manifold
@Student: I'm afraid with that topology you would not get a manifold structure on $TM$. Note that the topology you define is not Hausdorff, because if $p\in U$ and $x,y\in \pi^{-1}(p)$, then every open neighborhood of $x$ contains $y$ and vice versa.
May
4
comment Limit conditions of a subharmonic function imply that it is constant
@DanielS.: There is a form of the three circles theorem for subharmonic functions, which is that the function $m(r)$ is a convex function of $\log r$. For a reference, see for instance Hormander's Notions of Convexity Corollary 3.2.22, or just google "three circles theorem subharmonic". To derive the statement you are used to seeing about holomorphic functions, all you have to do is use that for a holomorphic function $f$, we have $\log|f|$ is subharmonic.
Mar
30
accepted Computing the degree of a finite morphism $\mathbb{P}^n\to \mathbb{P}^n$