Functional Analysis HW 2

Transcription

1 Brandon Behring Functional Analysis HW 2 Exercise 2.6 The space C[a, b] equipped with the L norm defined by f = b a f(x) dx is incomplete. If f n f with respect to the sup-norm then f n f with respect to. However, the converse does not hold. Proof: To show that C[a, b] equipped with L is incomplete, take any sequences of continuous functions that converges to the Heaviside unit step function in the L norm. Since step functions are not continuous, this would prove that our space is incomplete. For example, take the following functions on C[, ] if x < n n f n (x) = 2 (x + n ) if n x < n if n x and f(x) = { if x < if x We can compute the integra by looking at the areas in the graph to find that f f n L [,] = + f f n dx = 2 n. This shows that convergence in the L norm does not imply convergence in the sup-norm on C[, ] as f f n L [,] but f f n = 2 for all n. To show that if f n f with respect to the sup-norm then f n f with respect to, we use the mean value theorem for integrals (as both f and f n are continuous) to show that for some c [a, b] we have f f n = b a f(x) f n (x) dx = f(c) f n (c) (b a) sup f(y) f n (y) (b a) = f f n (b a). y [a,b]

2 Thus, if f n f with respect to the sup-norm then f n f with respect to L norm. Exercise 2.9 Let w : [, ] R be a nonnegative, continuous function. For f C([, ]), we define the weighted supremum norm by f w = sup {w(x) f(x) }. x if w(x) > for < x <, then w defines norm on C([, ]) and if w(x) > for x this norm is equivalent to the usual sup-norm. However, if w(x) = x, x is not equivalent to the usual sup-norm. Solution: We first show that w defines a norm. (i) f w for all f C([, ]). Proof: It is clear from f w = sup {w(x) f(x) }, x that f w is a supremum of nonnegative numbers as w(x) > so clearly f w. (ii) λf w = λ f w for all f C([, ]) and λ R. Proof: A direct computation gives λf w = sup {w(x) λ f(x) } x = sup {w(x) λ f(x) } x = λ sup {w(x) f(x) } x = λ f w. (iii) f + g w f w + g w for all f, g C([, ]). Proof: Preceding as we do in the case for the sup-norm f + g w = sup {w(x) f(x) + g(x) } x sup {w(x) f(x) + w(x) g(x) } x sup {w(x) f(x) } + sup {w(y) g(y) } x y = f w + g w. 2

3 (iv) f w = implies that f(x) = for all x [, ]. Proof: If f w = sup x {w(x) f(x) } = then we must have w(x) f(x) = for all x [, ]. Since w(x) > for all x [, ], then f(x) = for all in x [, ]. This final step fails if there exists x [, ] such that w(x) =. Proposition: w is equivalent to the usual sup-norm. Proof: We need to show that there exists m > and M > such that m f f w M f for all f C[, ]. Assuming w(x) is continuous (this is not explicitly stated in the problem, but I believe it was implied), we have from Theorem.68 that w(x) attains its minimum and maximum m and M for some values x, y [, ]. Let x [, ] be a value where we achieve a minimum, w(x) = m, and y [, ] be such a value where it achieves a maximum, w(y) = M. As w(x) > for all x [, ] clearly m > and M > since these values are obtained by values in [, ]. Thus, for all x [, ] we have that and thus < m < f(x) < M < sup {m f(x) } sup {w(x) f(x) } sup {M f(x) } x x x which implies what we wanted m f f w M f. This proof clearly breaks down if w(x) = for any x [, ] because we can no longer find an m >. In fact- it isn t even true that w is even a norm as we no longer even have that f w implies f(x) =. For an explicit example, consider f n (x) = ( x) n. This does not converge to a continuous function in the sup norm- it is one at zero and zero otherwise; thus f n (x) = for all n. However, f n (x) w(x)=x. Exercise 2.3 Consider the scalar initial value problem u(t) = u(t) α, u() =. Then the solution is unique if α but not if α <. Proof: We will examine four cases α >, α =, < α < and α =. 3

4 (i) If α =, then we have du dt = and u() =. The only solution is then u(t) = t. I believe there is a typo in the problem and that we have that is non unique for < α <. (ii) If α =, then we have du dt = u(t) with u() =. In the context of theorem 2.26, we need to show that f(u, t) = u is bounded in a rectangle R around the origin and is a Lipschitz continuous function of u, uniformly in t. The latter of course being trivial as we have no explicit t dependence. The Lipschitz condition is also clear from (See also Example 2.9) f(t, u) f(t, v) = u v u v. (iii) If α >, then we must show that f(u, t) = u α is bounded in a rectangle around the origin and is a Lipschitz continuous function of u. Let the rectangle R be contained in [ A, A] [ A, A]. Then f(u, t) = u α < A α so f(u, t) is bounded. Noting that { f αu( c) α if c < (u) = +αuc α if c. and α > gives us f (u) αa α for all u [ R, R]. The bounded derivative and the mean value theorem together gives us Lipschitz continuity as f(u, t) f(v, t) = f (c) u v αa α u v for all (u, t), (v, t) [ A, A] [ A, A]. Theorem 2.26 now implies a unique solution. Since u(t) = solves the ODE- it must be the only solution. (iv) If < α < we will follow Example 2.23 (α = 2 ) to construct a general counterexample. Solving the ODE f = f α f α df = dt α f(t) α = t C f(t) = ( α) (t C) α suggests we try { if t C u(t) = ( α) (t C) α if t > C. We now have a whole class of functions that solves our ODE as C is an arbitrary positive number so the solution is not unique. 4

5 Exercise 3.6 The following integral equation for f : [, a] R arise in a model of the motion of gas particles on a line: f(x) = + π a f(y)dy for a x a + (x y) 2 This equation has a nonnegative unique bounded, continuous solution for every a such that < a <. Proof: This is a Fredholm integral equation of the second kind for for f : [, a] R with g(x) = and k(x, y) = π By Theorem 3.3 we must show that sup { x [,a] By direct calculation we have + (x y) 2. k(x, y) dy} <. k(x, y) dy = π a x + (y x) 2 dy = π x + u 2 dy = ( tan (a x) tan ( x) ) = ( tan (a x) + tan (a + x) ). π π Additionally, because the range of tan (x) is we have {y R π 2 < y < π 2 }, ( sup tan (a x) + tan (a + x) ) < π. x [,a] Together, we have what we need sup { x [,a] k(x, y) dy} <. To see that the solution is always positive, we examine the Neumann series with g = f = ( K) = K n. 5 n=

6 As this is an infinite sum of positive values, it must be positive. If a is not finite, then if is a solution then f(x+c) = + π is also a solution. f(x) = + π + (x y) 2 f(y)dy + ([x + C] y) 2 f(y)dy = + π + (x u) 2 f(u+c)dy Exercise 3.7 There is a unique solution of the following nonlinear boundary value problem when the constant λ is sufficiently small u + λ sin u = f(x), u() =, u() =. Here, f : [, ] R is a given continuous function. Proof: Let us follow the reasoning set out between Proposition 3.4 and Theorem 3.5 in the text. However, instead of replacing f by qv + f, we now replace f with λ sin u + f to arrive at v(x) = λ sin(v(y))g(x, y)dy + which has the form (I K)v = h where g(x, y)f(y)dy Kv(x) = λ h(x) = sin(v(y))g(x, y)dy g(x, y)f(y)dy. We are now able to utilize Theorem 3.3, the Contraction Theorem applied to Integral Equations, if we can show that sup {λ x [,] sin(v(y))g(x, y) dy} < 6

7 for sufficiently small λ. Let A = sup { x [,] g(x, y) dy} (I believe the text claims that A = 8, but I can t verify this- either way it is clear that it is a finite number as ) then if λ < /A and noting that sin(x) for all x R we have sup {λ x [,] sin(v(y))g(x, y) dy} λ sup {λ x [,] λa <. g(x, y) dy} Let v =, then v (x) = λ sin()g(x, y)dy + g(x, y)f(y)dy = g(x, y)f(y)dy which could be in theory calculated if we knew f(y). Next, we have v 2 (x) = ( ) λ sin g(x, z)f(z)dz g(x, y)dy + g(x, y)f(y)dy Here- we see the non-linear integral equation becomes quite ugly to calculate by iteration- even though we do know v n will converge uniformly to our solution. 7