Solutions

a) We know that $\|cf\|$ must exist since $\{|cf(x)|:x\in[0,1]\}$ is bounded. For contradiction, suppose that $|c|\|f\|\ne\|cf\|$ for some $c\ne 0$.

If $|c|\|f\|<\|cf\|$, there must exist $x\in[0,1]$ such that

$$\|cf\|-\left(\|cf\|-|c|\|f\|\right)<|cf(x)|\le \|cf\|.$$

Since $\|cf\|$ is the least upper bound of $\{|cf(x)|:x\in[0,1]\}$, this means that

$$|c|\|f\|<|c||f(x)| \implies \|f\|<|f(x)|,$$

which cannot be because $\|f\|$ is an upper bound of $|f(y)|$ for $y\in[0,1]$.

Likewise, if $|c|\|f\|>\|cf\|$, then $\|f\|>\frac{1}{|c|}\|cf\|$, and there must exist $x\in[0,1]$ such that

$$\|f\|-\left(\|f\|-\frac{1}{|c|}\|cf\|\right)<|f(x)|\le\|f\|.$$

This implies

$$\|cf\|<|cf(x)|\le |c|\|f\|,$$

which invalidates $\|cf\|$ as an upper bound. Hence, it ensues that $\|cf\|=|c|\|f\|$ for all $c$. We left out the proof that $\|cf\|=|c|\|f\|$ for $c=0$ because it is bafflingly trivial.

b) Since $\|f+g\|$ is a least upper bound, for any $\varepsilon>0$ there exists $x\in[0,1]$ such that

$$\|f+g\|-\varepsilon<|f(x)+g(x)|\le \|f+g\|.$$

This implies that

$$\|f+g\|<\varepsilon+|f(x)+g(x)|\le \varepsilon+|f(x)|+|g(x)|\le \varepsilon+\|f\|+\|g\|.$$

Hence,

$$\|f+g\|\le \|f\|+\|g\|.$$

c) This inequality follows from b) by defining $s=h-g$ and $p=g-f$; we then have

$$\|s+p\|\le \|s\|+\|p\|.$$