Solutions

a) We will prove this by cases. If $f=0$, then

$$\max(f,0)+\min(f,0)=\max(0,0)+\min(0,0)=0+0=0.$$

Penultimately, if $f>0$, then

$$\max(f,0)+\min(f,0)=f+0=f.$$

Finally, if $f<0$, then

$$\max(f,0)+\min(f,0)=0+f=f.$$

b) For every $(x,f(x))\in f$, we define

$$\Delta(x)=|f(x)|+\varepsilon$$

for $\varepsilon\ge 0$. Let

$$g(x)=f(x)+\Delta(x)$$

and

$$h(x)=\Delta(x).$$

We have that both $g$ and $h$ are nonnegative, and we can write $f=g-h$ for infinitely many $\varepsilon$.