Marking scheme

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Question 1 : 10

Question 2 : (a) 5 (b) 5

Question 3 : 10

Question 4 :  (a) 2 (b) 2 (c) 3 (d) 3

Question 5 : (a) 5 (b) 5

Question 6 : (a) 2 (b) 3 (c) 2 (d) 3

Question 7 : (d) 4 (e) 3 (f) 3

Question 8 : (a) 2 (b) 3 (c) 2 (d) 2

Question 9 : (a) 5 (b) 5



* Question 1 :

Most people showed that T_g(f) does not depends on the representative of the equivalence class of f, but did not check that it also does not depend on the representative of the equivalence class of g.



* Question 2 :

Nothing to say, most people did everything right.  I liked when people used that equivalent norms means having the same topology, and continuity is a topological property so (a) is immediate. This avoids re-doing a proof that was essentially done already in the previous homework (with the identity mapping).


* Question 3 :  Very good overall.

One has to be careful with the word "bounded" the terms "T(U) is bounded" and "T|_U is bounded" are different. One cannot directly say "T(U) is bounded so T is continuous on U", for 2 reasons :

- Bounded has a different meaning, T(U) is a set (||Tu|| < M for u in U), it does not mean the operator is bounded (||Tu||< M||u|| for u in U). It is true that if U is an open set containing 0 and T(U) is bounded, then the function is continuous, but this is the exact thing you have to show in this question.

- The bounded operator = continuous operator words on normed vector spaces, here U is not  a vector space.


* Question 4 :

I did not remove points, but there is something almost     no one pointed out : the operator in (d) is indeed invertible, but no one checked that the inverse is well defined in the space, or in other words, that our function is surjective. If the map was defined over \ell^\infty or c_0, then the map (x_n) -> (nx_n) is not well defined.

For infinite dimensional spaces, having fg=id does not mean that f,g are invertible, one also needs gf=id (counter-example with left/right shifts of sequences.


Question 6 :

A lot did not attempt (d), or did not mention the fact that one of the two spaces is Banach.


Question 7 :

For (d) a lot tried to take the diagonal sequence and did the following subtle mistake doing so :

We know that x^m is Cauchy so indeed we have N such that for all m1,m2>N we have

||x^{m1} = x^{m2} ||<epsilon/2. This means that there is N' such that for all n>N' we have

||x^{m1}_n-x^{m2}_n || < epsilon. So far everything is correct BUT then people "take m1,m2,n > max(N,N') but N' depends on m1 and m2, so sometimes when one write ||x^{m1}_{m1}-x^{m2}_{m2} ||<epsilon, this is wrong, we have no control over them.


Question 8 :

For (c) a lot went a more complicated route than necessary, they said that if f^{-1}((-epsilon,epsilon)) is compact, then f is bounded, and do a few useless computations, before not using any of that and getting to the point that the compact set is bounded, so outside some ball, the function is small.