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Explicit twice differentiable bumps detect nonzero values of continuous functions

Statement

For real $c$ and $r > 0$ define $h : R \to R$ by $h (s) : = 10 s^{3} - 15 s^{4} + 6 s^{5}$ and $β_{c, r} : R \to R$ by

$β_{c, r} (t) : = h (1 - \frac{∣ t - c ∣}{r}) for ∣ t - c ∣ < r, β_{c, r} (t) : = 0 for ∣ t - c ∣ \geq r .$

Then:

$h (0) = 0$ , $h (1) = 1$ , $h^{'} (0) = h^{'} (1) = 0$ , $h^{''} (0) = h^{''} (1) = 0$ , and $0 \leq h \leq 1$ on $[0, 1]$ with $h (s) > 0$ for $0 < s < 1$ ;
$β_{c, r}$ is of class $C^{2}$ on $R$ ( $C^{k}$ Euclidean maps and diffeomorphisms), $0 \leq β_{c, r} \leq 1$ , $β_{c, r} (c) = 1$ , $β_{c, r} (t) > 0$ for $∣ t - c ∣ < r$ , and $β_{c, r} (t) = 0$ for $∣ t - c ∣ \geq r$ ;
(detection) let $J \subseteq R$ be an open interval, let $E : J \to R$ be continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point) and let $t_{0} \in J$ with $E (t_{0}) \neq 0$ . Then there is $ρ > 0$ with $[t_{0} - ρ, t_{0} + ρ] \subseteq J$ such that, with $β : = β_{t_{0}, ρ}$ ,

$\int_{t_{0} - ρ}^{t_{0} + ρ} E (t) β (t) d t \neq 0.$

Conventions. The displayed integral is the Darboux integral over the closed bounded interval $[t_{0} - ρ, t_{0} + ρ]$ of the continuous function $t \mapsto E (t) β (t)$ (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion); note that $β$ vanishes at both endpoints and outside the interval, so no extension convention is needed.

Facts & Assumptions

Given: The reals $c, r$ , the polynomials and bumps displayed in the Statement, an open interval $J$ , a continuous $E : J \to R$ and $t_{0} \in J$ with $E (t_{0}) \neq 0$ .

[F1]

$h$ is a polynomial, hence differentiable on $R$ ; the product and sum rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ) give $h^{'} (s) = 30 s^{2} (1 - s)^{2}, h^{''} (s) = 60 s (1 - s) (1 - 2 s),$ so $h (0) = 0, h (1) = 1, h^{'} (0) = h^{'} (1) = h^{''} (0) = h^{''} (1) = 0$ , and $h^{'} \geq 0$ on $R$ ; in particular $h$ is nondecreasing with $h (s) \in [0, 1]$ for $s \in [0, 1]$ and $h (s) > 0$ for $0 < s < 1$ .

[F2]

At an interior point of $R$ differentiability is the usual two-sided derivative, and at the points $c - r, c, c + r$ the one-sided derivatives of the two polynomial pieces of $β_{c, r}$ are computed by the chain rule (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ); the explicit matching argument below establishes two-sided differentiability at each gluing point.

[F3]

If $E$ is continuous at $t_{0}$ with $E (t_{0}) \neq 0$ , then for $ε = ∣ E (t_{0}) ∣ / 2 > 0$ there is $δ > 0$ such that $∣ E (t) - E (t_{0}) ∣ < ε$ for $∣ t - t_{0} ∣ < δ$ ; then $E (t) > E (t_{0}) / 2 > 0$ when $E (t_{0}) > 0$ and $E (t) < E (t_{0}) / 2 < 0$ when $E (t_{0}) < 0$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Absolute value in an ordered field). Since $J$ is open there is $ρ > 0$ with $[t_{0} - ρ, t_{0} + ρ] \subseteq J$ ; take $ρ \leq δ$ .

[F4]

A continuous real function on a closed bounded interval is integrable there (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion); the integral is linear and monotone (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ), and a nonnegative continuous function whose integral vanishes is identically zero (A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ ).

Proof

technique · direct; compute the polynomial and glue the pieces of the bump, then localise the sign of $E$ and integrate

1.1F1algebra

The derivative formulas of [F1] follow from the sum and product rules, and substituting $s = 0$ and $s = 1$ gives the vanishing values. Since $h^{'} \geq 0$ on $R$ , [F1] and the monotonicity criterion make $h$ nondecreasing on $R$ ; hence $0 = h (0) \leq h (s) \leq h (1) = 1$ for $s \in [0, 1]$ , and $h (s) > 0$ for $0 < s < 1$ because $h (s) = s^{3} (10 - 15 s + 6 s^{2})$ and $10 - 15 s + 6 s^{2} = 6 (s - 5 / 4)^{2} + 5 / 8 > 0$ . This proves clause 1.

1.2F1F2algebra

Clause 2: on each of the three regions $∣ t - c ∣ < r$ split at $t = c$ , and $∣ t - c ∣ \geq r$ , the function $β_{c, r}$ is a polynomial in $t$ composed with the affine map $t \mapsto 1 - (t - c) / r$ or $t \mapsto 1 + (t - c) / r$ , or is the constant $0$ ; by [F2] and the chain rule its one-sided derivatives of order $\leq 2$ at the gluing points $c - r, c, c + r$ are obtained from $h (0) = 0$ , $h (1) = 1$ , $h^{'} (0) = h^{'} (1) = 0$ , $h^{''} (0) = h^{''} (1) = 0$ of clause 1, and they agree from the two sides (at $c$ the factor $\mp 1 / r$ from the two affine branches multiplies $h^{'} (1) = 0$ , and at $c \pm r$ it multiplies $h^{'} (0) = 0$ , so the first derivative is $0$ from both sides; the second derivatives are $h^{''} (0) / r^{2} = 0$ and $h^{''} (1) / r^{2} = 0$ ). At each joining point $u$ the polynomial values agree, so the glued function is continuous. Its two one-sided difference quotients tend to the common first derivative $0$ , hence its two-sided derivative exists at $u$ . On either side this derivative is the polynomial derivative and has limit $0$ , so it is continuous there. Applying the same difference-quotient argument to this first derivative, whose one-sided derivatives both tend to $0$ , shows that the second derivative exists at $u$ and equals $0$ . Its polynomial pieces tend to that same value, making it continuous. Hence $β_{c, r}$ is of class $C^{2}$ on $R$ ; the bounds $0 \leq β_{c, r} \leq 1$ , the value $β_{c, r} (c) = h (1) = 1$ , the strict positivity on $(c - r, c + r)$ and the support statement are read off from clause 1 and the definition.

2.1F3step 1.2

Clause 3: by [F3] choose $ρ > 0$ with $[t_{0} - ρ, t_{0} + ρ] \subseteq J$ and with $E$ of the strict sign of $E (t_{0})$ on $(t_{0} - ρ, t_{0} + ρ)$ . Put $β : = β_{t_{0}, ρ}$ . On the support interval, $E β$ is continuous and has the strict sign of $E (t_{0})$ wherever $β > 0$ , and $E (t_{0}) β (t_{0}) = E (t_{0}) \neq 0$ ; by clause 2, $β > 0$ exactly on $(t_{0} - ρ, t_{0} + ρ)$ and $β \geq 0$ everywhere.

3.1F4step 2.1algebra∎

If $E (t_{0}) > 0$ then $E β \geq 0$ on $[t_{0} - ρ, t_{0} + ρ]$ and $(E β) (t_{0}) > 0$ ; were $\int_{t_{0} - ρ}^{t_{0} + ρ} E β = 0$ , [F4] would force $E β = 0$ identically on the interval, contradicting $(E β) (t_{0}) > 0$ . If $E (t_{0}) < 0$ apply the same argument to $- E β$ , whose integral is $- \int E β$ . In both cases the displayed integral is nonzero.

Remarks

The bump of this lemma is a polynomial, so no exponential is needed and the construction is entirely elementary; the price of using $C^{2}$ bumps instead of smooth ones is that the variations in The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient and Stationarity of the phase-space action is equivalent to Hamilton's equations, and hyperregular elimination recovers the Lagrangian action are taken of class $C^{2}$ , which is exactly the regularity required there.
The detecting clause is the local form of the fundamental lemma of the calculus of variations: it is used in Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression and Stationarity of the phase-space action is equivalent to Hamilton's equations, and hyperregular elimination recovers the Lagrangian action to show that a continuous coefficient which is not identically zero produces a nonzero first variation.

Explicit twice differentiable bumps detect nonzero values of continuous functions

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources