Question: Let $X$ be a Banach space and consider a bounded operator $A:X\to \mathbb R$.

> 
> I want to prove that $\|A\| = \alpha$ for a fixed $\alpha>0$ and I
>  already proved $\|A\|\leq \alpha$. So now ideally I would like to find
>  an $f$ such that $|Af| = \alpha$.
> 
> 
> 

I didn't quite manage to do that but instead I found a sequence $(f\_n)\_n$ such that $\|f\_n\|=1$ and $\lim\_{n\to \infty} |Af\_n| = \alpha$. Is this sufficient to conclude the claim?

Answer: 
Yes, you are correct, by Green's Theorem, you should evaluate
$$\int\_{x=0}^1\int\_{y=0}^{1-x}(y^2-x)dydx=\int\_{x=0}^1\left[\frac{y^3}{3}-xy\right]\_{y=0}^{1-x}dx=\int\_{x=0}^1\left(\frac{(1-x)^3}{3}-x(1-x)\right)dx\\
=\frac{1}{3}\int\_{t=0}^1t^3 dt-\int\_{x=0}^1 x(1-x)dx.$$
Can you take it from here?
$$\int\_{y=0}^1\int\_{x=0}^{1-y}(y^2-x)dxdy = \int\_{x=0}^1(y^2x-\frac{x^2}{2})|\_{0}^{(1-y)}dy$$

If you expand you get
$$=\frac{1}{2}\int\_{y=0}^1(-2y^3+y^2+2y-1)dy$$

$$=\frac{1}{2}\left(\frac{-2y^4}{4}+\frac{y^3}{3}+\frac{2y^2}{2}-y\right)|\_{0}^{1}$$

$$=-\frac{1}{12}$$