Home Topics in Analysis Analysis Elementary Linear Algebra Linear Algebra Analysis of Functions of Many Variables Linear Algebra and Analysis Complex Analysis Calculus of One And Many Variables Engineering Math One Variable Advanced Calculus Interrogation of Hiroshi Oshima

126 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES

Now if n is still larger, continuity of f shows that | f (x)− pn (xn)|< 4ε. Since ε is arbitrary,pn (xn)→ f (x) and so, passing to the limit with this subsequence in 5.3 yields 1.

Now consider 2. It holds for polynomials p(x) obviously. So let ∥pn− f∥→ 0. Then∫ c

apndx+

∫ b

cpndx =

∫ b

apndx

Pass to a limit as n→ ∞ and use the definition to get 2. Also note that∫ b

b f (x)dx = 0follows from the definition.

Next consider 3. Let h ̸= 0 and let x be in the open interval determined by a and b.Then for small h, F(x+h)−F(x)

h = 1h∫ x+h

x f (t)dt = f (xh) where xh is between x and x+ h.Let h→ 0. By continuity of f , it follows that the limit of the right side exists and so

limh→0

F (x+h)−F (x)h

= limh→0

f (xh) = f (x)

If x is either end point, the argument is the same except you have to pay attention to thesign of h so that both x and x+h are in [a,b]. Thus F is continuous on [a,b] and F ′ existson (a,b) so if G is an antiderivative,∫ b

af (t)dt ≡ F (b) = F (b)−F (a) = G(b)−G(a)

The claim that the integral is linear is obvious from this. Indeed, if F ′ = f ,G′ = g,∫ b

a(α f (t)+βg(t))dt = αF (b)+βG(b)− (αF (a)+βG(a))

= α (F (b)−F (a))+β (G(b)−G(a))

= α

∫ b

af (t)dt +β

∫ b

ag(t)dt

If f ≥ 0, then the mean value theorem implies that for some

t ∈ (a,b) ,F (b)−F (a) =∫ b

af dx = f (t)(b−a)≥ 0.

Thus∫ b

a (| f |− f )dx ≥ 0,∫ b

a (| f |+ f )dx ≥ 0 and so∫ b

a | f |dx ≥∫ b

a f dx and∫ b

a | f |dx ≥−∫ b

a f dx so this proves∣∣∣∫ b

a f dx∣∣∣≤ ∫ b

a | f |dx. This, along with part 2 implies the other claim

that∣∣∣∫ b

a f dx∣∣∣≤ ∣∣∣∫ b

a | f |dx∣∣∣.

The last claim is obvious because an antiderivative of 1 is F (x) = x. ■Note also that the usual change of variables theorem is available because if F ′ = f , then

f (g(x))g′ (x) = ddx F (g(x)) so that, from the above proposition, F (g(b))− F (g(a)) =∫ g(b)

g(a) f (y)dy =∫ b

a f (g(x))g′ (x)dx.We usually let y = g(x) and dy = g′ (x)dx and thenchange the limits as indicated above, equivalently we massage the expression to look likethe above. Integration by parts also follows from differentiation rules.

Definition 5.8.6 If f ∈ Cc (R) , define∫R f ≡

∫∞

−∞f (x)dx as

∫ ba f (x)dx where the

interval [a,b] is chosen such that spt( f )⊆ [a,b].

126 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACESNow if 7 is still larger, continuity of f shows that | f (x) — pp (%n)| < 4e. Since é is arbitrary,Pn (Xn) 2 f (x) and so, passing to the limit with this subsequence in 5.3 yields 1.Now consider 2. It holds for polynomials p (x) obviously. So let ||p, — f|| + 0. Thenc b b/ prdx+ [ Pndx = / PndxJa Je JaPass to a limit as n + and use the definition to get 2. Also note that iP f (x)dx =0follows from the definition.Next consider 3. Let h # 0 and let x be in the open interval determined by a and b.Then for small h, Forth)“ FG) = 1 pet! ¢ (1) dt = f (xn) where x, is between x and x +h.Let h — 0. By continuity of f, it follows that the limit of the right side exists and sokm F (x +h) — F (x)h—0 h= lim f (xs) = F(x)If x is either end point, the argument is the same except you have to pay attention to thesign of h so that both x and x+/ are in [a,b]. Thus F is continuous on [a,b] and F’ existson (a,b) so if G is an antiderivative,[roaero — F (b) —F (a) =G(b)—G(a)The claim that the integral is linear is obvious from this. Indeed, if F’ = f,G' = g,[er +Betyar = aF (b)+BG(b)—(aF (a) + BG(a))= a(F(b)—F(a))+B(G(b)—Ga))= af riyar+B [ gtyarIf f > 0, then the mean value theorem implies that for someb1 € (a,b) ,F (b) —F (a) = / fdx = f(t) (b—a) >0.JaThus fi (|f|—f)dx > 0, Je (|f|+f)dx > 0 and so fq |fldx > fi fax and f? |fldx >— ? fdx so this proves | [? fdx| < {?|f|dx. This, along with part 2 implies the other claimthat |? fax] < | ff |fldxThe last claim is obvious because an antiderivative of | is F (x) =x. liNote also that the usual change of variables theorem is available because if F’ = f, thenf (g(x)) 8! (x) = 4F(g (x)) so that, from the above proposition, F (g(b)) — F (g(a)) =fe £0) dy = {? f (g(x))g! (x) dx.We usually let y = g(x) and dy = g' (x)dx and thenchange the limits as indicated above, equivalently we massage the expression to look likethe above. Integration by parts also follows from differentiation rules.Definition 5.8.6 if f € C.(R), define fx f = [@.f (x) dx as [° f (x)dx where theinterval [a, b| is chosen such that spt (f) C |[a, }).