I looked up the answer for this online but couldn't find one that was satisfactory and used the hint provided. Here's the question
Let $h(z)$ be a continuous complex-valued function on the unit interval $[0,1]$, and consider $$H(z) = \int_0^1 \frac{ h(t)}{t - z} dt.$$ Where is $H(z)$ defined? Where is $H(z)$ continuous? Justify your answer. Hint: Use the fact that if $|f(t) - g(t)| < \epsilon$ for $0 \leq t \leq 1$, then $\int_0^1 |f(t) - g(t)| dt < \epsilon$.
I realize that $H(z)$ should be defined and continuous for any $z \notin [0,1]$, but I'm not sure how to formally prove it. Thanks
We estimate $$ |H(z+\eta)-H(z)|=\left |\int_0^1 h(t)\frac{\eta}{(t-z-\eta)(t-z)}\mathrm dt \right|\\ \leq |\eta|M\int_0^1 \frac{1}{|(t-z-\eta)(t-z)|}\mathrm dt $$ where $M=\sup_{t\in [0,1]}|h(t)|$, finite by the extreme value theorem.
Now, fix $z\in \mathbb{C}\setminus [0,1]$. Then, define $$ f:[0,1]\to \mathbb{R}\\ x\mapsto |x-z| $$ a continuous function on a compact set, and thus achives it's minimum, call it $\delta>0$. Now, picking $|\eta|<\delta/2$ we get $$ |t-z-\eta|\geq ||t-z|-|\eta||=|t-z|-|\eta|\geq \delta/2 $$ then using the fact in the hint and that our estimates hold for any $t\in[0,1]$ we have $$ |H(z+\eta)-H(z)|\leq |\eta|M\frac{1}{\delta(\delta/2)} $$ Now it is clear that picking $|\eta|<\delta_*=\min\left\{ \frac{2\epsilon}{M\delta^2},\delta/2\right\}$ suffices to meet any $\epsilon$ challenge.