 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, $|f+g|\le |f|+|g|$. Each is nonnegative. If $\Vert f{\Vert }_{\infty }=0$, 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 $\Vert f{\Vert }_1>0$. Thus both are norms.

For the nonnegative $f_n$, direct integration gives the L1 normTherefore $\boxed{f_n⟶0\text{ in }\Vert \cdot {\Vert }_1}$. On the other hand, differentiation places the maximum of $t^n(1-t)$ at $t=n/(n+1)$, giving the supremum normFor $t<1$, exponential decay implies $n{t}^n(1-t)\to 0$, 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 there is no convergence in the supremum norm.