Prove Holomorphic on C: Continuity and Differentiability

[fauboca]

Yes, something like that would work.

So I am trying to use Morera's Theorem:
Let U be an open set in C and let f be continuous on U. Assume that the integral of f along the boundary of every closed rectangle in U is 0. Then f is holomorphic.

So let U = \mathbb{C} - [2,5] Let R be rectangles in U which are parallel to the coordinate axes. So \int_{\partial R}f=0.

Now how can I use this?

 

[HallsofIvy]

By the way, "holomorphic at every point" doesn't make sense- "holomorphic" is not a point property. You may mean "analytic at every point". "Holomorphic" means "analytic at every point in the complex plane".

 

[fauboca]

I have this proof for finite points but how would I modify it for infinite many points between [2,5]?

Assume q(z) is any function that is holomorphic on a disc U except at a finite number of points \xi_1,\ldots, \xi_n\in U, and assume \lim_{z\to\xi_j}(z-\xi_j)q(z)=0 for 1\leq j\leq n. Let U'=U-\{\xi_1,\ldots\xi_n\}. Then q is holomorphic on U'.
Note F(z)=q(z)-f'(a) so q(z)=\frac{f(z)-f(a)}{z-a}

Step 1 is the Cauchy-Goursat argument:
\int_{\partial R}q(z)dz=0 for all rectangles R in U such that \xi_j\notin\partial R for j.

proof:

Let R be a rectangle with the boundary of R in U'. First subdivide R into sub-rectangles so that each sub-rectangle has at most one \xi_j inside it. By Cauchy-Goursat, \int_{\partial R}=\sum_i\int_{\partial R_i}. So it suffices to show \int_{\partial R}q=0 if R contains at most one \xi_j.
If R contains no \xi_j, then we are done by Cauchy-Goursat. Assume \xi=\xi_j is inside R. Let \epsilon>0 be given. Put \xi in a square of size x at the center of this where x is chosen small enough so that |(z-\xi)q(z)|<\epsilon for z in and on this square. Subdivide this rectangle so the square which is x by x is its own rectangle around \xi. As before, the integrals of the sub-rectangles that don't contain \xi are 0. So \int_{\partial R}q=\int_{\text{square with xi}}q.
$$
\left|\int_{\text{square with xi}}q\right|\leq \underbrace{||q||_{\text{square with xi}}}_{\frac{\epsilon}{\min\{|z-\xi|\}}}\times \underbrace{(\text{length of square with xi}}_{4x}\leq\frac{\epsilon}{x/2}4x=8\epsilon
$$
So \left|\int_{\text{square with xi}}q\right|=0.

Step 2 is to use step one to create a primitive for q on all of U.
Define g(a)=\int_{z_0}^{z_1}q(z)dz where the path is from z_0 horizontal and then vertical. So g(a) is well-defined. If a point is not unreachable, then \lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=q(a) by exactly the same means as before. Unreachable are the points vertical above \xi.
When computing \frac{g(a+h)-g(a)}{h}, only consider h in C with |h|<\frac{\delta}{2}. The path for computing g(a+h) also misses \xi. Then for these h g(a+h)-g(a)=\int_a^{a+h}q so the same reason as before show \frac{g(a+h)-g(a)}{h}-g(a)\to 0 as h\to 0.

To handle all the bad exception.
Pick \epsilon\geq 0 such that the point z_1=z_0+\epsilon(1+i) is not on any of the same vertical or horizontal lines as any \xi_j. Of course epsilon is really small compared to the radius of U.
Define g_1(a)=\int_{z_1}^aq(z)dz. This defines a primitive for q(z) on all of the disc except on the unreachable regions.
Define g_2(a)=\int_{z_1}^aq(z)dz but this time move vertical and then horizontal. So g_2(a) is also a primitive for q on U except at the its unreachable points (but g_1 reaches those points and vice versa). On the overlap, g_1 and g_2 are primitives for the same q. They differ only by a constant. But g_1(z_1)=0=g_2(z_1) so g_1=g_2 on the overlap. For a small enough epsilon,
$$
g(z)=\begin{cases}g_1(z)\\g_2(z)\end{cases}
$$ is defined on all of U' and is a primitive for q.

How do I extend this to my problem?

 

[morphism]

By the way, "holomorphic at every point" doesn't make sense- "holomorphic" is not a point property. You may mean "analytic at every point". "Holomorphic" means "analytic at every point in the complex plane".

Not true; what you're thinking of is "entire".

re: fauboca

Here's what you do. Let $\gamma$ be a closed curve in C. If $\gamma$ doesn't contain any point of [2,5] in its interior, then $\int_\gamma f =0$ since f is holomorphic away from [2,5]. So now suppose that $\gamma$ contains points of [2,5] in its interior. For the sake of illustration, suppose that $\gamma$ is a square with vertices (3,0), (3,1), (4,1) and (4,0). Cut the square horizontally into two rectangles of length 1-ε and ε, so that the smaller rectangle has its base on [3,4]. This breaks up your contour into two contours, and the integral of f over one of them is zero. Now use a limit argument.

The general case is similar.

 

[fauboca]

Not true; what you're thinking of is "entire".

re: fauboca

Here's what you do. Let $\gamma$ be a closed curve in C. If $\gamma$ doesn't contain any point of [2,5] in its interior, then $\int_\gamma f =0$ since f is holomorphic away from [2,5]. So now suppose that $\gamma$ contains points of [2,5] in its interior. For the sake of illustration, suppose that $\gamma$ is a square with vertices (3,0), (3,1), (4,1) and (4,0). Cut the square horizontally into two rectangles of length 1-ε and ε, so that the smaller rectangle has its base on [3,4]. This breaks up your contour into two contours, and the integral of f over one of them is zero. Now use a limit argument.

The general case is similar.

Shouldn't the square contain the interval? So it would be (1.9,-.1), (1.9, .1), (5.1,-.1),(5.1,.1)?

 

[fauboca]

Let $\gamma$ be a closed curve in $\mathbb{C}$. If $\gamma$ doesn't contain any point from [2,5] in its interior, then $\int_{\gamma}f=0$ since f is holomorphic away from [2,5]. Suppose that $\gamma$ contains [2,5] in its interior. Let R be a rectangle oriented with the coordinate axes in $\gamma$ such that [2,5] is in R such that $[2,5]\notin\partial R$. Divide the rectangle into two sub rectangles of length $1-\epsilon$ and $\epsilon$ such that [2,5] is contained in one the sub rectangles. WLOG suppose [2,5] is the rectangle of length $\epsilon$. Then the over f of the rectangle of length $1-\epsilon$ is zero.

I am not sure how you mean to use the limit argument.