Hilbert-space operator norm in L2 [closed]

The first part is to compute $\left\lVert A(f)\right\rVert$: $$\left\lVert A(f)\right\rVert_2^2=\left\lVert f-\int_0^1 f(x)dx\right\rVert_2^2=\int_0^1\left(f(y)-\int_0^1 f(x)dx\right)^2dy=$$$$=\int_0^1\left(f(y)^2-2f(y)\int_0^1 f(x)dx+\left(\int_0^1 f(x)dx\right)^2\right)dy=$$$$=\int_0^1 f(y)^2dy-2\int_0^1f(y)dy\int_0^1 f(x)dx+\left(\int_0^1 f(x)dx\right)^2=$$$$=\int_0^1 f(y)^2dy-\left(\int_0^1 f(y)dy\right)^2$$ Then: $$\left\lVert A(f)\right\rVert_2^2\leq \int_0^1 f(y)^2dy=\left\lVert f\right\rVert_2^2$$ So $\left\lVert A\right\rVert\leq 1$. The second part is to find an $f$ such that the inequality becomes an equality, in this case the inequality is given by the factor $\left(\int_0^1 f(y)dy\right)^2$, so we need an $f$ such that this integral is $0$, for example $f=x-\frac{1}{2}$:

$$\left\lVert A(f)\right\rVert_2^2=\int_0^1 f(y)^2dy=\left\lVert f\right\rVert_2^2\Rightarrow \left\lVert A(f)\right\rVert_2=\left\lVert f\right\rVert_2$$ Then $\left\lVert A\right\rVert= 1$.

In general the aproach for these kind of problems is always the same: find a bound $\left\lVert A(f)\right\rVert_2\leq M\left\lVert f\right\rVert_2$ and then find an $f$ such that the inequality becomes an equality, if you can't this is probably because the $M$ is not good enough. Then $\left\lVert A\right\rVert=M$ and you are done.