# Independently analytic and continuous, but not jointly continuous?

In Bak/Newman's "Complex Analysis", they write:

17.9 Theorem
Suppose $\phi(z,t)$ is a continuous function of $t$, with $b \ge t \ge a$, for fixed $z$ and an analytic function of $z \in D$ for fixed $t$. Then
$$f(z) = \int _a ^b \phi(z,t) \ dt$$ is analytic in $D$, and ...etc...

The proof starts off:

Since $f$ is a continuous function of $z$, according to Morera's Theorem we need only prove that ...etc...

I cannot seem to force $f$ to be continuous without requiring $\phi$ to be continuous in both variables together. I feel like it might be that analytic in the first variable and independently continuous in the second does not imply jointly continuous.

Question: What, if there is one, is an example which is analytic in the first, continuous in the second, but not jointly continuous? Any reasonable $\text{(domain) }D \times [a,b]$ is OK.

EDIT: Theorem 5.4 on p. 56 of Stein/Shakarchi's book here seems to be pretty much the same, except with the joint continuity assumption. (Their proof is neat, too, because it avoids blatantly using Fubini's theorem).

Thank you!

It seems that you have to assume joint continuity, it does not follow from the assumptions. First, by Runge's theorem there exists a sequence of polynomials $$p_n$$ such that $$p_n \to 0$$ pointwise in the unit disk, but not uniformly in any neighborhood of $$0$$. For details see Pointwise convergence of sequences of holomorphic functions to holomorphic functions. Then you can define the function $$\phi: \mathbb{D} \times [0,1] \to \mathbb{C}$$ by $$\phi(z,0) = 0$$, $$\phi(z,1/n) = p_n(z)$$, and extend this piecewise linearly in $$t$$. The function $$\phi$$ is analytic in $$z$$ for every $$t$$, continuous in $$t$$ for every $$z$$, but it is not continuous at $$(0,0)$$. (If it were, then $$p_n(z) = \phi(z,1/n)$$ would converge uniformly to $$0$$ on every compact subset of the unit disk.)
Note that this construction does not guarantee that the function $$f$$ defined in the question is discontinuous, but I am pretty sure the Runge construction is flexible enough to adjust the construction in order to produce discontinuous $$f$$.
• @reuns: I am not extending the function piecewise constantly, but piecewise linearly. For every fixed $z$, the function $\phi(z,t)$ is continuous on $\{0,1,1/2,1/3,\ldots\}$, since $p_n(z) \to 0$. Then I just extend this by linear interpolation to each interval $(1/(n+1),1/n)$. And I don't quite understand your second question, there is no assumption of uniform boundedness. If one adds that assumption, Cauchy's estimates for the derivative and some standard complex analysis arguments would imply continuity of $\phi$, that is true. – Lukas Geyer May 3 '19 at 20:47
• Ok, then what about $q_n(z)$ the Runge polynomial uniform $\epsilon=1/n$ approximation to $z\log z$ on $1/n^2\le |z|\le 2, arg(z) \in [0,2\pi-1/n^2]$ and $p_n(z) = q_n(z e^{2i\pi /n}+1/n)$ ? This way with $\phi(z,t) = \int_0^{1/10} p_{\lfloor 1/(t+u) \rfloor}(z) du$ then $\int_0^1 \phi(z,t)dt$ should be holomorphic only on $0<|z| < 1,arg(z) \in (0,2\pi)$ – reuns May 3 '19 at 23:15
• @reuns: As I said, the discontinuity of $\phi$ does not guarantee the discontinuity of $f(z) = \int \phi(z,t) \, dt$, but the Runge construction is very flexible, and I am sure with a bit more work one can construct such an example for which the integral is discontinuous at zero. – Lukas Geyer May 3 '19 at 23:23
• @reuns: Also, I expect that by Osgood's theorem the function $f$ can not be too bad, because pointwise limits of analytic functions are still locally uniform on an open dense set, so one would probably expect $f$ to be analytic on an open dense set (though I do not have a proof of that.) – Lukas Geyer May 4 '19 at 0:08