13.1. SARD’S LEMMA AND APPROXIMATION 303

be in ∂K. Therefore, K = K1 but this also is a contradiction because if x ∈ ∂K then x /∈ Kthanks to the first part that ∂U =U \U . ■

Note that for an open set U ⊆ Rp, and h : U → Rp, dist(h(∂U) ,y)≥ dist(h(U),y)

because U ⊇ ∂U .The following lemma will be nice to keep in mind.

Lemma 13.0.4 f ∈C(Ω× [a,b] ;Rp

)if and only if

t→ f (·, t) ∈C([a,b] ;C

(Ω;Rp))

Also∥f∥

∞,Ω×[a,b] = maxt∈[a,b]

(∥f (·,t)∥

∞,Ω

)Proof:⇒By uniform continuity, if ε > 0 there is δ > 0 such that if |t− s|< δ , then for

all x ∈Ω, ∥f (x,t)−f (x,s)∥< ε

2 . It follows that

∥f (·, t)−f (·,s)∥∞≤ ε

2< ε

⇐Say (xn, tn)→ (x,t) . Does it follow that f (xn, tn)→ f (x,t)?

∥f (xn, tn)−f (x,t)∥ ≤ ∥f (xn, tn)−f (xn, t)∥+∥f (xn, t)−f (x, t)∥≤ ∥f (·, tn)−f (·, t)∥

∞+∥f (xn, t)−f (x, t)∥

both terms converge to 0, the first because f is continuous into C(Ω;Rp

)and the second

because x→ f (x, t) is continuous.The claim about the norms is next. Let (x, t) be such that ∥f∥

∞,Ω×[a,b] < ∥f (x, t)∥+ε .Then

∥f∥∞,Ω×[a,b] < ∥f (x, t)∥+ ε ≤ max

t∈[a,b]

(∥f (·, t)∥

∞,Ω

)+ ε

and so ∥f∥∞,Ω×[a,b] ≤ maxt∈[a,b] max

(∥f (·,t)∥

∞,Ω

)because ε is arbitrary. However, the

same argument works in the other direction. There exists t such that

∥f (·, t)∥∞,Ω = max

t∈[a,b]

(∥f (·, t)∥

∞,Ω

)by compactness of the interval. Then by compactness of Ω, there is x such that

∥f (·,t)∥∞,Ω = ∥f (x, t)∥ ≤ ∥f∥

∞,Ω×[a,b]

and so the two norms are the same. ■

13.1 Sard’s Lemma and ApproximationFirst are easy assertions about approximation of continuous functions with smooth ones.

The following is the Weierstrass approximation theorem. It is Corollary 5.5.3 presentedearlier.