It is frustrating that maximize works only numerically (inside plot).

Not even after

F:=simplify(eval(f,t=x*m));

maximize(F, x=0..2*Pi);
or
maximize(F, x=-infinity..infinity);

produces a syntax error instead of the correct result:

@Carl Love 

You can't hope for an answer to such an elliptic question!
And what is sol?
I don't think that you will find mind readers in this forum.

@AmirHosein Sadeghimanesh 

Please provide an example.

@Markiyan Hirnyk 

For a>0, k>0 the the expression under sqrt at the numerator is <0. Now, indeed, the principal branch of sqrt is discontinuous at any z < 0, but the restriction of sqrt to (-oo, 0] is continuous, and this is enough for the continuity of our function. That is why I said that the continuity follows easily.

A variant of Joe Riel's solution also works:

L:= [1,2,3, {4,5,6}]

subsindets(L, {set,list}, op);

@Bendesarts 

@Preben Alsholm 

Probably I was wrong, it's a very old version I seem to recall.
Anyway, the main reason for the alias is to not use a(t) at all.

@Markiyan Hirnyk 

When I say easy, I really mean easy, at least for an undergraduate level.
We are both professional mathematicians, as I understand.
Anyway, I am going to answer in the future only to polite questions, i.e. without "Roly-poly toy", "empty words" etc.

restart:
alias(a=a(t)):
a - a(t);   ###  it used to work!
   a - a(t)

@John Fredsted 

You are right here. The image of the connected set (0, oo) ^ 4 by a continuous function is connected, so the function  must be constant since the only possible values are I and -I. The continuity is easy but not automatic, because the principal branch of sqrt is not continuous in C.

Preben's solution is correct, the function w being C^1 in the convex open domain {(a,k) : a>0, k>0}.

OK, you win. It was my mistake to answer your "objections" following the pattern:

Assertion: p>0, q>0 implies p*q>0.

Objections:

1. Empty words.
2. Unbased words.
3. Prove it.
4. Prove it with Maple.
5. Refresh your logic.
6. Go to 1.

@Markiyan Hirnyk 

@Markiyan Hirnyk 

But t is >0; you had no objections about this. Now you come back? Infinite loop?
As I see you want to test my patience.

@Markiyan Hirnyk 

I have mentioned that this follows from ex^2=-1.

is( (- sqrt(p)* t )^2  < 0  )  assuming p>=0, t>0;

    false

is( (- sqrt(p)* t )^2  < 0  )  assuming p<0, t>0;

    true

@Markiyan Hirnyk 

argument( - sqrt(p)* t ) assuming p<0, t>0;

@Markiyan Hirnyk 

If you indicate which assertion is not obvious, I will try to turn my empty words into full ones.