Isometrically isomorphic normed spaces

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Let X and Y be normed spaces. If X and Y are isometrically isomorphic, then their duals X' and Y' are also isometrically isomorphic.

I have no idea what to do with this, please help. :(
 
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This is just a matter of writing down what you have, and understanding the definitions. Say you have an isometric isomorphism F:X->Y. We want to get an isometric isomorphism G:X'->Y'. The question you should be asking yourself is: If we start with an element f in X', how can we use this to get an element G(f) in Y'?
 
Duality if contravariant. Surely a morphism F:X->Y more naturally (no pun) gives a map G:Y'->X'.
 
Right. I had initially written my G as going from Y' to X', but decided to stick with X'->Y' in case the OP doesn't immediately see why we've switched directions. But regardless, the moment he/she sets up the obvious map, he/she will probably notice that it's easier to go from Y' to X'.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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