 Solution

The integral is finite and nonnegative because $|f|$ is continuous on a compact interval. If it vanishes and $f(x_{\ast })\ne 0$, continuity gives a relative interval of positive length on which $|f|\ge |f(x_{\ast })|/2$, a contradiction. Thus the integral is zero only for $f=0$. For a real scalar $a$, $\Vert af\Vert =|a|\Vert f\Vert$, and the pointwise triangle inequality $|f+g|\le |f|+|g|$ integrates to $\Vert f+g\Vert \le \Vert f\Vert +\Vert g\Vert$. These are all the norm axioms.

The given sequence satisfiesHence it converges to the zero function in this integral norm. Although $f_n(0)=1$ for all $n$, an isolated endpoint contributes no integral. The pointwise limit is discontinuous, and the sequence does not converge in the uniform norm; neither fact obstructs convergence in the stated norm.