= Solution
Both functionals are finite on $X$: a <continuous function> on the <compact metric space> $[0,1]$ is bounded and integrable. Homogeneity follows from $|af|=|a||f|$, and the <triangle inequality> follows by integrating, or taking the supremum of, $|f+g|\leq|f|+|g|$. Each is nonnegative. If $\|f\|_\infty=0$, then $f$ vanishes everywhere. If a <continuous function> $f$ is nonzero at some point, continuity supplies an interval of positive length on which $|f|$ is bounded below by a positive number; hence $\|f\|_1>0$. Thus both are <norms>.
For the nonnegative $f_n$, direct integration gives the <L1 norm>
$$
\|f_n\|_1=n\int_0^1(t^n-t^{n+1})\,dt=\frac{n}{(n+1)(n+2)}\longrightarrow0.
$$
Therefore $\boxed{f_n\longrightarrow0\text{ in }\|\cdot\|_1}$. On the other hand, differentiation places the maximum of $t^n(1-t)$ at $t=n/(n+1)$, giving the <supremum norm>
$$
\|f_n\|_\infty=\left(\frac{n}{n+1}\right)^{n+1}\longrightarrow e^{-1}.
$$
For $t<1$, exponential decay implies $nt^n(1-t)\to0$, while $f_n(1)=0$. Thus the <pointwise convergence> is to zero throughout $[0,1]$. Any convergence in the <supremum norm> would imply <uniform convergence>, hence the same pointwise limit zero, contradicting the displayed positive limiting <norm>. Therefore \b[there is no convergence in the supremum <norm>].
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