Is it possible that $||K||= 1$ but $||K^2|| = 0$ with $||K|| = \max_{0 \leq x \leq 1} \int_0^1|k(x,y)|dy$ The question is 

Define $K:C([0,1]) \to C([0,1])$ by 
  $$
Kf(x)= \int_0^1k(x,y)f(y)dy 
$$
  where $k:[0,1] \times [0,1] \to \mathbb{R}$ continuous. 
  
  
*
  
*Prove that $K$ is bounded and 
  $$
||K|| = \max_{0 \leq x \leq 1} \int_0^1|k(x,y)|dy
$$
  
*Is it possible that $||K||= 1$ but $||K^2|| = 0$

I did finish the first part, but I don't know how to do the second one. In particular, how to approach $||K^2|| = 0$? Thanks.
 A: We compute using Fubini:
$$ K^2 f(x) = \int_0^1  k(x,y) Kf(y) dy
= \int_0^1 k(x,y) \int_0^1 k(y,z) f(z) dz dy
= \int_0^1 f(z) \Big( \int_0^1 k(x,y) k(y,z) dy \Big) dz $$
If $\Vert K^2 \Vert =0$, then we have for all $f\in C^0([0,1])$ and all $x\in [0,1]$
$$0 = K^2 f(x) = \int_0^1 f(z) \Big( \int_0^1 k(x,y) k(y,z) dy \Big) dz $$
As we can pick (fix $x$ for a second)
$$ f(z) =\int_0^1 k(x,y) k(y,z) dy $$
we obtain that for all $x,z \in [0,1]$ must hold
$$ \int_0^1 k(x,y) k(y,z) dy =0.$$
Hence, we get
$$ \Vert K^2 \Vert = 0 \quad \Leftrightarrow \quad \forall x\in [0,1] \forall z\in [0,1]: \int_0^1 k(x,y) k(y,z) dy =0 $$
Now let us make some happy guessing. What would make our life easier, well only one variable would be so much easier. We make the ansatz $k(x,y)=g(x)$. Then we get
$$ \int_0^1 k(x,y) k(y,z) dy = \int_0^1 g(x) g(y) dy = g(x) \int_0^1 g(y) dy $$
Hence, we can just pick some continuous $g$ with $\int_0^1 g(y)dy =0$. Now we might be worried that $\Vert K \Vert \neq 1$. Not a problem. Pick $g$ as above, not being the constant zero function (I leave it to you to write down such a function). Then the following function will do the trick
$$ k(x,y) = \frac{g(x)}{\max_{z\in [0,1]} \vert g(z) \vert} $$
I leave it to you to check that his indeed gives a function with the properties you want.
