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Statement in the source convention

Fix an integer $n\ge2$ and distinct nodes in the descending order used by Erdős and Turán,

$$1\ge x_1>x_2>\cdots>x_n\ge-1.$$

Put

$$\omega(x)=\prod_{j=1}^n(x-x_j), \qquad l_j(x)=\frac{\omega(x)}{\omega'(x_j)(x-x_j)} =\prod_{i\ne j}\frac{x-x_i}{x_j-x_i}.$$

The product is the polynomial interpretation of the displayed quotient at $x=x_j$. Then, for every $1\le k\le n-1$,

$$\boxed{ l_k(x)+l_{k+1}(x)\ge1 \quad\text{for every }x\in[x_{k+1},x_k]. } \tag{ET-IV}$$

This is Lemma IV on printed p. 529 / physical p. 20 of Erdős--Turán (1940). The paper first writes the proof for $2<k<n-2$ and then says that $k=1,2,n-2,n-1$ are analogous. The argument below retains its derivative-root count and spells out all endpoint cases.

Literal source issue. The final sentence on printed p. 529 visibly says that an interior point with $\phi(\xi_0)=1$ would force three derivative zeros. The desired lower bound requires exclusion of $\phi(\xi)<1$. The proof below does not silently change the printed equality sign: after reproducing the source's endpoint-derivative and root-budget method, it gives the short mean-value argument that excludes a value below one. That step is a completion written here, not an author-issued correction.

Proof

Write

$$\phi(x)=l_k(x)+l_{k+1}(x).$$

The cardinal identities give

$$\phi(x_k)=\phi(x_{k+1})=1, \qquad \phi(x_j)=0\quad(j\notin\{k,k+1\}), \tag{1}$$

and $\deg\phi\le n-1$. For $n=2$, the polynomial $\phi-1$ has degree at most one and vanishes at both nodes, so $\phi\equiv1$. Assume henceforth that $n\ge3$. In particular, $\phi'$ is not identically zero, because $\phi$ also has at least one zero in (1), and

$$\deg\phi'\le n-2. \tag{2}$$

We first handle $2\le k\le n-2$, so both exterior neighbors $x_{k-1}$ and $x_{k+2}$ exist. Rolle's theorem applied between consecutive zero nodes on the two sides of $[x_{k+1},x_k]$ gives distinct zeros of $\phi'$ in

$$(x_{j+1},x_j), \quad 1\le j\le k-2 \quad\text{or}\quad k+2\le j\le n-1. \tag{3}$$

Empty index ranges contribute nothing. There are

$$(k-2)+(n-k-2)=n-4 \tag{4}$$

such zeros, all outside the central interval. Equation (1) and Rolle's theorem supply another zero in $(x_{k+1},x_k)$.

We claim that the endpoint derivatives point into a possible arch above the level one:

$$\phi'(x_k)\le0, \qquad \phi'(x_{k+1})\ge0. \tag{5}$$

Suppose first that $\phi'(x_k)>0$. Since $\phi(x_k)=1$ and $\phi(x_{k-1})=0$, the mean-value theorem gives a point of $(x_k,x_{k-1})$ where $\phi'<0$. Continuity then gives a zero of $\phi'$ in that interval.

If $\phi'(x_{k+1})\ge0$ as well, the equality of the two values in (1), together with $\phi'(x_k)>0$, forces $\phi'$ to be negative somewhere in $(x_{k+1},x_k)$. Otherwise its integral across that interval would be positive. Continuity would then give two distinct zeros of $\phi'$ in the central interval. Together with (3) and the zero in $(x_k,x_{k-1})$, this would give at least

$$(n-4)+2+1=n-1$$

distinct zeros of $\phi'$, contradicting (2). Hence $\phi'(x_{k+1})<0$. But $\phi(x_{k+2})=0$ and $\phi(x_{k+1})=1$, so the mean-value theorem gives a point of $(x_{k+2},x_{k+1})$ where $\phi'>0$. There is consequently a zero between that point and $x_{k+1}$. Counting this zero, the zero in $(x_k,x_{k-1})$, the central Rolle zero, and the $n-4$ zeros in (3) again gives $n-1$ distinct zeros, a contradiction. Thus $\phi'(x_k)\le0$. Reversing left and right proves $\phi'(x_{k+1})\ge0$, establishing (5).

If (ET-IV) failed, there would be a $\xi\in(x_{k+1},x_k)$ with $\phi(\xi)<1$. The mean-value theorem would give points

$$u\in(x_{k+1},\xi),\qquad v\in(\xi,x_k)$$

with $\phi'(u)<0<\phi'(v)$. In conjunction with (5), continuity would give three distinct zeros of $\phi'$: one in $[x_{k+1},u)$, one in $(u,v)$, and one in $(v,x_k]$. Adding the $n-4$ exterior zeros in (3) gives at least $n-1$ distinct zeros, again contradicting (2). This proves (ET-IV) for $2\le k\le n-2$.

It remains to make the source's endpoint phrase explicit. If $k=1$, then $\phi$ vanishes at $x_3,\ldots,x_n$, so Rolle's theorem gives $n-3$ zeros of $\phi'$ in the intervening open intervals. It gives one more in $(x_2,x_1)$ because $\phi(x_2)=\phi(x_1)=1$. If $\phi'(x_2)<0$, the change from $\phi(x_3)=0$ to $\phi(x_2)=1$ first gives a positive derivative and hence an additional zero in $(x_3,x_2)$, exceeding the degree bound (2). Thus $\phi'(x_2)\ge0$. If $\phi'(x_1)>0$, equality of the endpoint values forces a negative derivative in $(x_2,x_1)$ and hence two zeros there; together with the $n-3$ exterior zeros, this also exceeds (2). Thus $\phi'(x_1)\le0$. A hypothetical value $\phi(\xi)<1$ would now give three central zeros exactly as above, again contradicting (2). This proves the case $k=1$. Reflecting the nodes and reversing their indices proves $k=n-1$. All indices and all $n\ge2$ are therefore covered. $\square$

Increasing-node form and the 1978 interface

For increasing nodes $-1\le y_1<\cdots<y_n\le1$, let $L_j$ be their fundamental polynomials. Relabeling $x_i=y_{n+1-i}$ in (ET-IV) gives

$$L_j(y)+L_{j+1}(y)\ge1 \qquad (y_j\le y\le y_{j+1}), \quad 1\le j\le n-1. \tag{6}$$

Both terms in (6) are nonnegative. This follows directly from their product form: on $[y_j,y_{j+1}]$, every numerator factor in $L_j$ or $L_{j+1}$ has the same sign as its corresponding denominator factor. Consequently

$$|L_j(y)|+|L_{j+1}(y)| =L_j(y)+L_{j+1}(y)\ge1. \tag{7}$$

Equation (6) is the form in which P. Erdős and J. Szabados cite Lemma IV on p. 194 of their 1978 paper on the integral of the Lebesgue function ([[polynomials/erdos_szabados_1978_integral_lebesgue_function_interpolation/integral_lower_bound|its integral lower bound]]); Erdős's 1989 paper on convergent interpolatory polynomials also uses the lemma ([[polynomials/erdos_1989_convergent_interpolatory_polynomials/theorem|its theorem]]). Those applications are not reviewed here.

Dependencies. The definition and cardinal identities for ordinary Lagrange fundamental polynomials; Rolle's theorem, the mean-value theorem, the intermediate-value theorem, the fundamental theorem of calculus, and the fact that a nonzero real polynomial of degree at most $d$ has at most $d$ distinct real zeros. No theorem from the preceding 19 pages of the 1940 article enters this proof.

Source qualification. Complete rewritten proof of Lemma IV from printed p. 529 / physical p. 20, with the paper's terse analogous endpoint cases, its literal final equality sentence, the strict-inequality completion, and the increasing-node relabeling made explicit. The rewritten proof has an independent review, retained as the Lemma IV review; it does not review the rest of Erdős--Turán (1940).

Source. P. Erdős and P. Turán, On interpolation. III. Interpolatory theory of polynomials, Annals of Mathematics (2) 41 (3) (1940), 510--553, DOI 10.2307/1968733; Lemma IV and its proof on p. 529.

Bears on. Problem 1153: read through (7), the lemma gives only $|l_k(x)|+|l_{k+1}(x)|\ge1$ between adjacent nodes, hence $\lambda(x)\ge1$, and no bound that grows with $n$. It is an input that Erdős and Szabados (1978) cite on p. 194 for their integral lower bound, whose own relation to the problem is stated on [[polynomials/erdos_szabados_1978_integral_lebesgue_function_interpolation/integral_lower_bound|that page]]; the lemma supplies no constant for the problem.