How to “normalize” an age-of-death probability distribution with respect to births in corresponding years?

Let's suppose that we know about the age of death of n people (this being a representative sample of a given population), all died in the same year H.

We can define $P(x)$, the probability of dying at each age $x$ in the year H, as the percentage of people in the sample dead at the age of $x$: $$P(x)=\frac{m(x)}{n}$$ where $m(x)$ is the number of people in the sample dead at the age of $x$.

Let's say that this distribution looks something like this:

At this point one can see that even if the curve is globally quite "smooth", there do are a few local irregolarities (like little "hills" and "valleys") and so a first operation to correct this behaviour could be to "normalize" the generic value $P(\bar{x})$ with respect to the births in the corresponding year H-$\bar{x}$ (all these birth data, from year H-$x_{-}$ to year H-$x_{+}$, are supposed to be well known).

What could it be the best mathematical method to do so?

I tought of multiplying each generic value $P(\bar{x})$ by the fraction $$\frac{b.a.}{b(H-\bar{x})}$$ where $b.a.$ is the mean number of births per year, from year H-$x_{-}$ to year H-$x_{+}$, and $b(H-\bar{x})$ is the number of births in the year $H-\bar{x}$; re-normalizing after that so that all the $P(x)$s sums to one again.

Still, this procedure only seems to give the curve more asperities.

Any help will be much appreciated, thank you in advance!

The graph that you obtained is actually a histogram, rather than a curve, since it reports values for integer values of age at death. Data irregularities in histograms can often be "smoothed" by reducing the class width (e.g., in this case considering intervals lower than one year on the $x$-axis), which often requires to increase the sample size, and/or by resampling (when possible). Where additional sampling are impossible to acquire (as probably in this case), a solution is to try smoothing techniques, for example the 'moving average" or others.