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.
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||
Below is an instruction that describes a task. Write a response that appropriately completes the request.

### Instruction:

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.

### Response:

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||
