Approximation and Complexity, I

Approximation and Complexity, I

December 9, 2019

1. Let

P

= {v ∈ P

: v(2) = 0}.

Find coefficients a

of the polynomial v(x) =

a

x

∈ P

that best approximates the function

f (x) = x

− x

− x − 1 in the uniform norm on [−1, 1].

Solution

The space P

is a 3-dmiensional Haar space, so that we can use the Chebyshev alternation theorem. Since f (2) = 1, the optimal polynomial v satisfies the equation

f (x) − v(x) = T

(x) T

(2) ,

where the alternation points are cos(

) for 0 ¬ j ¬ 3. Since T

(x) = 4x

− 3x, we have T

(2) = 26 and

v(x) = f (x) − T

(x)

T

(2) = (x

− x

− x − 1) −

(4x

− 3x) =

x

− x

−

x − 1.

2. Let f ∈ C([0, 1]) and g ∈ C

(R) be correspondingly defined as f (x) = 4x

and g(t) = 2(1 + cos t) − sin

t.

Show that

dist(f, P

) = dist(g, V

).

Solution

Since g is an even function, its best approximation in V

is also even and spanned by 1 and cos t. We have

g(t) = 1 + 2 cos t + cos

t = 4



1 + cos t 2

2

= 4x

= f (x),

where x =

(1 + cos t), the transformation [0, π] 3 t 7→ x ∈ [0, 1] being one-to-one.

For any polynomial x 7→ a

+ a

x from P

and any x ∈ [0, 1] we have

f (x) − (a

+ a

x) = g(t) −

a

+

a

(1 + cos t)

= g(t) −

(a

+

a

) +

cos t

. And vice-versa; for any polynomial t 7→ a

+ a

cos t we have

g(t) − (a

+ a

cos t) = f (x) −

a

+ a

(2x − 1)

= f (x) −

(a

− a

) + 2a

x)

. Hence dist(f, P

) and dist(g, V

) are equal.

1

3. Let n ­ 1. For f ∈ C([0, 1]), let p

be the polynomial that best approximates f in the uniform norm on [0, 1] among all algebraic polynomials p ∈ P

satisfying

p(0) = f (0) and p(1) = f (1).

Show that

kf − p

k ¬ 2 dist(f, P

).

Solution

Let p

be the optimal polynomial for f in P

. Let ` be the polynomial of degree at most 1 that interpolates the data `(0) = f (0) − p

(0) and `(1) = p

(1). Obviously

k`k = max

|f (0) − p

(0)|, |f (1) − p

(1)|

¬ kf − p

k ¬ dist(f, P

).

Then for p = p

+ ` we have p ∈ P

, p(0) = f (0), p(1) = f (1), and

kf − p

k ¬ kf − pk = kf − (p

+ `)k ¬ kf − p

k + k`k ¬ 2 dist(f, P

), as claimed.

4. Let x

= i/n for 0 ¬ i ¬ n. Let V

be the linear subspace of C([0, 1]) consisting of functions that are polynomials of degree ¬ 1 on each subinterval [x

, x

]. Define the operator L

: C([0, 1]) → V

in such a way that L

f is the element in V

that interpolates f at the points x

; that is,

(L

f )(x

) = f (x

) for 0 ¬ i ¬ n.

(a) Is L

a projection onto V

? (b) What is the norm of L

?

(c) Does L

f converge to f for all f ∈ C([0, 1])?

(d) If the answer to (c) is ‘yes’; can the convergence be arbitrarily slow?

Solution

(a) and (b): The operator L

is linear and onto V

, by definition. We also have that kL

f k = max{|f (x

|, . . . , |f (x

)|} ¬ kf k,

so that L

is bounded and kL

k ¬ 1. On the other hand, for f = 1 we have L

f = f so that kL

k = 1. Hence L

is a projection onto V

whose norm equals 1 for all n.

(c): Since the operators L

are positive, for convergence for all continuous functions f it suffices that we have convergence for the three functions h

(x) = x

, i = 0, 1, 2.

We obviously have L

h

− h

= 0 and L

h

− h

= 0. To estimate the error for h

we use the well-know error formula for polynomial interpolation. For x ∈ [x

, x

] we have

|L

h

(x) − h

(x)| = 1

2 |(x − x

)(x − x

)f

(ξ

)| = |(x − x

)(x − x

)| ¬ 1 4n

, which means that lim

kL

h

− h

k = 0.

(d): The convergence can be arbitrarily slow due to the Bernstein’s lethargy theorem

applied to the sequence of spaces V

⊂ V

⊂ V

⊂ · · · ⊂ V

⊂ · · · .