Question: Let E be a normed space and F a Banach space. Prove that the set of bounded linear operators from E to F (which you can assume to already be a vector space with respect to suitable operations and field), equiped with the operator norm, is also a Banach space.

Answer: To prove that the set of bounded linear operators from E to F, denoted by L(E, F), equipped with the operator norm, is a Banach space, we need to show that it is complete, i.e., every Cauchy sequence of operators in L(E, F) converges to an operator in L(E, F).

Let {Tn} be a Cauchy sequence of operators in L(E, F). This means that for every ε > 0, there exists N such that ||Tn - T_m|| < ε for all n, m > N. Our goal is to show that this sequence converges to some operator T in L(E, F).

First, we need to show that Tn converges pointwise to an operator T. Let x ∈ E, and fix ε > 0. Since {Tn} is Cauchy, there exists N such that ||Tn - Tm|| < ε for all n, m > N. Now, consider the inequality:

||Tn(x) - Tm(x)|| ≤ ||Tn - Tm|| * ||x|| < ε * ||x||.

This inequality implies that {Tn(x)} is a Cauchy sequence in F. Since F is a Banach space, it is complete, so Tn(x) converges to some element T(x) ∈ F.

Now, we need to show that T is linear and bounded. Linearity follows from the linearity of each T_n. To show that T is bounded, we need to show that there exists a constant C such that ||T(x)|| ≤ C * ||x|| for all x ∈ E.

Fix ε > 0, and let N be such that ||Tn - Tm|| < ε for all n, m > N. Then for any x ∈ E, we have:

||T(x) - TN(x)|| = ||lim(Tn(x)) - TN(x)|| = lim(||Tn(x) - T_N(x)||) ≤ lim(ε * ||x||) = ε * ||x||.

We can rearrange this inequality to get:

||T(x)|| ≤ ||T_N(x)|| + ε * ||x||### Human: Let *a* and *b* be positive integers such that  *ab + 1*  divides  *a2 + b2*. Prove that *(a2 + b2) / (ab + 1)* is the square of an integer.### Assistant: To prove that (a^2 + b^2) / (ab + 1) is the square of an integer, we need to show that it is an integer and that its square root is an integer.

First, note that (a^2 + b^2) / (ab + 1) is an integer if and only if (a^2 + b^2) - k(ab + 1) = 0 for some integer k. Rearranging, we get:

a^2 - kab + b^2 - k = 0

Viewing this as a quadratic equation in a, the discriminant must be a perfect square, that is:

k^2 - 4(b^2 - k) = m^2 for some integer m.

Simplifying, we get:

k^2 - 4b^2 + 4k - m^2 = 0

This is a quadratic equation in k, and its discriminant must also be a perfect square, that is:

16b^2 - 4m^2 = n^2 for some integer n.

This can be rewritten as:

(4b + n)(4b - n) = 4m^2

Since 4b + n and 4b - n are both even, we can write them as 2r and 2s, respectively, for some integers r and s. Then we have:

rs = m^2 / 4

and

b = (r + s) / 2

Substituting into the original equation, we get:

a^2 - (r + s)a + [(r - s) / 2]^2 = 0

This is a quadratic equation in a, and its discriminant must also be a perfect square, that is:

(r + s)^2 - 4[(r - s) / 2]^2 = t^2 for some integer t.

Simplifying, we get:

(r - s)^2 + t^2 = (r + s)^2

This is the Pythagorean theorem, and its solutions are:

r = uv, s = uw, t = u(v^2 - w^2)

where u, v, and w are positive integers with v > w and (v - w) odd.

Therefore, we have:

b = (r + s) / 2 = uwv

and

a = (r - s) / 2 = u(v^2 - w^2)

Thus, we have shown that (a^2 + b^2) / (ab + 1) is an integer, and its square root is:

(a^2 + b^2) / (ab + 1) = (u^2v^2 + u^2w^2) / u^2vw + 1 = (v^2 + w^2) / vw + 1 = (v/w)^2 + 1

which is clearly an integer. Therefore, (a^2 + b^2) / (ab + 1) is the square of an integer.