Say that a function $f(t)$ “changes only by jumps” if $f(t) + \text{const} = C ∑_k j_k θ(t-t_k)$ for a certain constant $C$. Here $θ(t)$ is the Heaviside step function which has a jump 1 at $t=0$ (it is 0 for $t<0$ and 1 for $t≥0$), and $f(t)$ has jumps $j_k$ at points $t_k$. (The reason for using a coefficient $C$ is shown below; in the simplest cases $C=1$.)

Obviously, this holds for piecewise-constant functions with finitely many jumps: the sum is finite, and it indeed has jumps $j_k/C$ at points $t_k$, with $C=1$. This may be restated as: “the actual jumps” of the sum coincide with “the prescribed jumps” $j_k$. The same happens when the series converges absolutely (in particular, the sum is right- and left-continuous).

In general, without the condition of absolute convergence “everything may happen”. However, what I would never expect is an example of “a reasonable function” $f(t)$ for which the identity above holds with $C=½$. In other words: the actual jumps of the sum are twice the prescribed jumps.

I discovered such an example when examining plots of certain Fourier series appearing¹⁾ in the Artin case of the Langlands program.
**Question:** are there other interesting cases when $C≠1$? Or other similar “pathologies”?

¹⁾ In fact, me considering this case was a result of misunderstanding on my part. (So I do not know whether this example is in the scope of the “real” Langlands program.) See my notes; search for all places mentioning Eisenstein.

This essentially ends my question. However, since the claim above about C=½ seems so incredible, below I provide the details of my
example. — And this takes **quite some space** (sorry)!

(An impatient reader may want to jump to the picture below showing plots of two functions with “reasonably defined” jumps at

disjointsubsets of ℚ.)

### Restating in terms of derivative

By the way: one can restate the question in terms of $\operatorname{d}f=f'(t)\operatorname{d}t$ — and then instead of $f$ “changing only by jumps” it is about “atomic” generalized functions, which are sums of shifts of atomic δ-measures. The advantage of this approach is that in our example $\operatorname{d}f$ is actually a boundary value of a modular 1-form, hence is a modular-invariant 1-form on ℝℙ¹.

My gut feeling is that appearance of “extra” coefficient 2 is due to “effects of a certain ‘hidden symmetry’”. One can notice that
in our example $\operatorname{d}f$ has not only the modular symmetries, but also the “functional equation” symmetry $t ↦ -1/13t$. (It is not modular
since there is no matrix in GL₂ℚ with determinant ±1 which matches this projective transformation.) Moreover, this symmetry
*exchanges* differentials of two parts of $f$ considered below.

However, I cannot see how this symmetry may be explicitly related to the “extra” coefficient 2. I suspect a presence of “yet another discrete symmetry”…

## Details of the example

The *prescribed* jumps are going to be $±1/Q$ at every (irreducible) rational $P/Q$. To explain the signs, recall that when working with
modular forms and modular symbols²⁾, the simplest examples are Eisenstein series. At a real value $t$ of the argument, they may be
written as $f(t) ≔ ∑_{r=0}^t J(r)$, with the summation running over rational numbers, and $J(P/Q)$ a certain function on ℚ. (Here and
below for simplicity we take $t≥0$. Of course, there is an immediate generalization for $t<0$.)

²⁾ … which are antiderivatives of modular forms. These may be a little bit easier to work with than “proper” modular forms; indeed the latter ones are defined on $\{\operatorname{Im} z > 0\}$ and are too singular near ℝ; hence their restrictions to ℝ are “only” generalized functions. On the other hand, taking the antiderivative “mollifies” singularities, hence the restrictions of the antiderivative may be “smooth enough” to “almost” have values at points. (At least, in the cases we consider here they have left and right limits.)

Note that $f'(t)$ is “atomic”, i.e., a sum of shifts $J(r)δ(t-r)$ of δ-functions at $r∈ℚ$. Compare to the reformulation above.

Obviously: if the sum converges absolutely, then $f(t)$ has right and left limits, and its jump at $t=r$ is $J(r)$. For example, this happens if $J(P/Q)=B(P/Q)/Q^s$ with $s>2$ and $B(r)$ bounded (for this formula to make sense, assume $P$ and $Q$ coprime and $Q>0$). However, in many cases³⁾ the sum can be assigned a “reasonable” sense even if there is no absolute convergence. For example, one can sum first over rationals with denominator 1, then with 2, etc. — since this specifies an order of summation, one can ask whether the series converges or not.

³⁾ For example, the sum described above is an analytic function of $s∈C$ for $\operatorname{Re}s>2$. It may have an analytic continuation to a larger region in ℂ. (Note that for many “summation methods” the result of summation remains an analytic function of $s$. Using a “stronger” summation method than “taking lim of partial sums” may extend the region of convergence. In other words, such a summation method would “expose a part of the analytic continuation of the sum”.)

Moreover, convergence means that in the sum, positive $J(r)$ “mostly compensate” negative $J(r)$: two sums over positives/negatives diverge, but taken together two speeds of divergence cancel each other. For example, if for fixed $Q$ one assigns $B(P/Q)$ randomly to be ±1 or 0 so that the frequency of $+1$ is the same as the frequency of $-1$, then $B(r)$ with the denominator $Q$ contribute (roughly speaking) $O(\sqrt Q)/Q^s$ into the sum. Hence the sum converges (almost surely) if $s>1$. (Such questions of probabilistic convergence are addressed by “Kolmogorov’s 2-series criterion”.)

In fact, if the cancellation in⁴⁾ $S(t,Q) ≔ ∑_{P=0}^{⌊tQ⌋} B(P/Q)$ is even stronger than for random $B$, then the sum may converge for $s=1$. And this is what happens for “Eisenstein series of weight 2 with quadratic character”.

⁴⁾ I abuse notations since the sum must skip $P$ not coprime with $Q$.

### Quadratic Eisenstein series of weight 2

For example, in the case when the conductor $c≡₄1$ is ±prime, $B(P/Q)$ is the Legendre “quadratic residue” symbol⁵⁾ $ℒ(Q \operatorname{mod} c)$ unless $c|Q$, otherwise $B(P/Q) = ℒ(P \operatorname{mod} c) \sqrt c$ (here we again assume that $P$,$Q$ are coprime, and $Q>0$).

⁵⁾ Abuse of notation: for negative $c$, we should take the Kronecker symbol, not the Legendre one. So in the example we take $c>0$.

This breaks the series into two parts (for $c∤Q$ and $c|Q$). Here are the plots of the partial sums⁶⁾ of these parts with $Q≤l$ (and $Q≤cl$ for the second part) with $l=1000$, $c=13$ over their period:

⁶⁾ In fact, I cheated a bit: the first partial sum is taken with a certain regularization. (Details below.)

One can see that the blue plot has the prescribed jumps at $P/Q$ with $13|Q$, and the red one behaves likewise with $13∤Q$. However,
*otherwise* the plots more or less coincide!

CONCLUSION: when $l→∞$, two parts of the sum coincide. Hence *every* part would have the jumps of *both* types. And the sum of these
two parts would have *twice* the prescribed jumps! This finishes our example of how $C$ may be equal to ½.

(Well, the argument above only *describes* what happens. The details and proofs are in the paper linked to above. However, this
is the case when having a proof does not bring any satisfaction — I still have no clue *why* this phenomenon appears…)

### The regularization

For completeness, I discuss a minor cheating used above.

Without regularization, the partial sum over $Q≤1000$ with $13∤Q$ would acquire a “parasitic” linear term $M_{1000}·t$ with $M_{1000}≈0.748$.
When increasing the limit of summation $l$, *on average* $M_l→0$. (E.g., this happens with “Cesàro summation” in $Q$ instead of a
“rigid cutoff after $Q=l$” used in the definition of a partial sum.) So instead of using a suitable fancy summation, we may just
correct every sum (in $P$) over a particular $Q$ by $±t·φ(Q)/Q$. (After this correction, this sum in $P$ becomes 1-periodic in $t$.)

To “show” that such correction has the same effect as “a regularized summation”, it is enough to show that a *suitable* regularized
sum of the coefficients $±t·φ(Q)/Q$ at $t$ vanishes. So let us *define* this sum via an analytic continuation (as above) by replacing $Q$
in the denominator by $Q^s$; then we need to show that the value at $s=1$ vanishes. Using Euler products, one gets:
$$ ∑_Q ℒ(Q,c)φ(Q)/Q^s = ∏_p (1/p + (1-1/p)/(1-ℒ(p \operatorname{mod}c)/p^{s-1})) = Φ(s-1)/Φ(s)\\
\text{ with }Φ(s) ≔ 1/∏_p(1-ℒ(p \operatorname{mod}c)/p^s) = ∑_nℒ(n \operatorname{mod}c)/n^s. $$
The manipulations make perfect sense for $\operatorname{Re}s>2$, when the convergence is absolute. Moreover, $Φ(s)$ converges for $\operatorname{Re}s>0$ with $Φ(1)≠0$;
hence if we can show that $\lim_{s→1+} Φ(s-1)=0$ this implies $Φ(s-1)/Φ(s) → 0$ when $s→1+$.

Indeed, $∑_{k=-v}^v ℒ(k \operatorname{mod}c)=0$; moreover, $∑_{k=-v}^v kℒ(k \operatorname{mod}c)=0$ since $c≡₄1$ (so $ℒ(k \operatorname{mod}c)$ is even in $k$); here $|c|=2v+1$. These sums appear when one plugs Taylor expansion in $k$ of $1/n^s$ with $n=cm+k$ into $Φ(s) = ∑_nℒ(n \operatorname{mod}c)/n^s$: $$ ∑_m∑_kℒ(k \operatorname{mod}c)·(1/(cm)^s - sk/(cm)^{s+1} + ℛ(m,k,s)). $$ Hence two first terms of the Taylor expansion $1/(cm)^s$ and $-sk/(cm)^{s+1}$ contribute 0 into this sum (when $m≥1$). What remains is the $n$⸣s not covered by our substitution $n=cm+k$ (so $1≤n≤v$), which sum to $½∑_{k=-v}^v ℒ(k \operatorname{mod}c)=0$ when $s=0$, and the remainders $ℛ(m,k,s)$ in the Taylor series which are $s(s+1)k²ℒ(n \operatorname{mod}c)/2(cm+Δ_{k,m})^{s+2}$ (with $|Δ_{k,m}|<v$). The only things about remainders important to us are the absolute convergence (for $\operatorname{Re}s>-1$, which covers a neighborhood of $s=0$), and the factor $s$ which shows that the value at $s=0$ vanishes.