Important results in the theory of PDEs regarding boundary-value problems are trace and extension theorems. Since the trace operator (not to be confused with the trace from linear algebra) essentially acts by restriction to the boundary of the domain, I was wondering how it got the name "trace": who came up with this name, when and why?

(I asked this earlier on MathOverflow but did not get an answer.)


2 Answers 2


Normally, I'd refer to Jeff Miller's Earliest Known Uses of Some of the Words of Mathematics site for these kinds of questions, but it is silent on this one, and generally "trace"'s history seems unexplored.

The first use of "trace" in the sense of restriction to the boundary on MathSciNet is (as far as I could find) in the review of Slobodeckii's 1958 note Sobolev's Spaces of Fractional Order and Their Application to Boundary Problems for Partial Differential Equations (Dokl. Akad. Nauk SSSR (N.S.) v. 118, 1958, pp. 243–246). The reviewer was Lions, who himself began to use the term afterwards, perhaps the first in the West, e.g. in the monograph Équations Différentielles Opérationnelles et Problèmes aux Limites (1961). The author of the first "trace theorems", Sobolev, does not use the word in his well-known book, Some Applications of Functional Analysis in Mathematical Physics (1950), which summarized his work from 1930-s and brought him the world fame. But it does occur in his Varenna lectures Sur les Equations aux Dérivées Partielles Hyperboliques Non-linéaires published in Rome in 1961 as a monograph:"The first four lectures develop the properties of generalized derivatives, including important theorems about the traces of functions on varieties in $R^n$ (in particular, the hyperplanes $t=\,$const". So it might have been Slobodeckii's doing, but I wouldn't claim it without more direct evidence.

As for the name's motivation, I can only speculate. "The traces of functions on varieties" is reminiscent of algebraic geometry, where intersections of surfaces with planes were traditionally called their "traces", see e.g. Lardner's 1831 Treatise on Algebraic Geometry, p.229. If one thinks of functions as graphs over the domain then graph's intersection with the cylinder over the boundary (in nice cases) will be exactly the "trace" in Slobodeckii's sense.


The name Trace here is related to the notion of the trace of function in general which basically means the restriction.
I other word, the Trace of a function in general its restriction (whenever it makes sense) on a subset of ITS DOMAIN ( usually named as trace set of the original function) .

But be aware the actual situation is much more delicate. Since given a function defined almost everywhere on a bounded domain it is not always possible to take its TRACE (or its restriction ) on the boundary.

Rather if we consider a CONTINUOUS function up the the boundary , it is possible to take its TRACE (or its restriction) on the boundary. This means that the OPERATOR that maps a continuous function up the boundary to it TRACE (to its restored) on the boundary is WELL define . Moreover, this OPERATOR is bounded if we endowe both depature and arrived spaces with suitable L.p. or Sobolev NORMS. THIS THEN GIVES THE NAMES TRACE OPERATOR.

FINALLY since the space of continuous function up to boundary is dense IN SOME SUITABLE L.p. space or SOBOLEV SPACE, we know that, OUR LATEST TRACE OPERATOR CAN BE CONTINUOUSLY EXTEND TO THE corresponding L.p. or Sobolev spaces.


Remark: 1. There are some cases where the density argument may not apply but One can still use HANH-BANACH theorem to get A CONTINUOUS EXTENTION. 2. One may chose other spaces instead of the space of continuous function up the boundary. Like the space consisting of functions which are restrictions of functions that are smooth (on a bigger domian ) on closure of the DOMIAN (the we are working with ).

  • 4
    $\begingroup$ The OP is not asking what the "trace" means or how it is defined, it is asking who first used the word for this purpose. There is no obvious passage from "restriction" to "trace" etymologically. $\endgroup$
    – Conifold
    Oct 17, 2017 at 23:46

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