# Limit of integral

Let $g: \mathbb{C}\times[a,b]\to\mathbb{R}$ a continuous function and

$$g(t)=g(h_0,t)=\lim\nolimits_{h\to h_0} g(h,t) \ \forall t\in \mathbb{R}$$

Is the following result/reasoning correct ? If we have a sequence of functions, we need uniform convergence to pass the limit inside the integral, but it seems it works everytime here. Or there is a flaw ?

We have

$$\lim_{h\to h_0} \int_a^b g(h,t)dt= \int_a^b g(t)dt$$

because

Idea 1:

$$\forall t\in[a,b], \ \forall > \epsilon >0, \exists \delta(t)>0\text{ > such that }|g(h,t)-g(t)|<\epsilon \ > \forall |h-h_0|<\delta(t)$$

Then if $|h-h_0|<\delta=\inf_{t\in[a,b]} > \delta(t)$,

$$|g(h,t)-g(t)|<\epsilon \ \forall > |h-h_0|<\delta(t), \ \forall > t\in[a,b]$$

Thus

$$\left|\int_a^b > (g(h,t)-g(t))dt\right|\le \int_a^b > |g(h,t)-g(t)|dt\le (b-a)\epsilon$$

Doesn't work if $\inf_{t\in[a,b]} > \delta(t)=0$, see George's comment.

~

Idea 2: (added after George's comment)

Let $\overline{B(h_0, \gamma)}\subset > C$ a closed ball. Resctrict $g$ to $> \overline{B(h_0, \gamma)}\times[a,b]$. Because it is continuous on a compact, this function will be uniformly continuous. Thus

$$\forall \epsilon >0 \ \exists > \delta>0 \ : \ > |g(h,t)-g(h_0,t)|<\epsilon \ \forall > |h-h_0|<\min(\delta, \gamma)$$

and

$$\left|\int_a^b > (g(h,t)-g(t))dt\right|\le \int_a^b > |g(h,t)-g(t)|dt\le (b-a)\epsilon$$

if $|h-h_0|<\min(\delta, \gamma)$.

Does this one work ;) ?

Thank you for any help :)

-
The proof would fall apart if $\inf_{t\in[a,b]}\delta(t)$ was zero. You want a single $\delta > 0$ to apply for all t, for which uniform continuity is needed. – George Lowther Nov 8 '10 at 18:54
Thank you! I felt there was a problem with this infimum, but I could not tell what. I found another way to proceed, and added it to the question. Does it fail as well ? – Klaus Nov 8 '10 at 19:58
Yes that works! I could add it as an answer, but it hardly seems worth it now that you have the correct answer in the post. – George Lowther Nov 8 '10 at 20:12
Nice! Thank you for your valuable help. – Klaus Nov 8 '10 at 20:15

So, the second idea was the right one: considering $g$ on a compact to have it uniformly continuous. See my message above.