= Solution
The <integral> is finite because a <continuous function> on a <compact> <closed interval> is bounded. Nonnegativity is immediate, and <absolute homogeneity of a norm> follows from $|cf(x)|=|c|\,|f(x)|$. The pointwise <triangle inequality> gives
$$
\|f+g\|\leq\int_{-1}^1(|f(x)|+|g(x)|)\,dx=\|f\|+\|g\|.
$$
For positive definiteness, if $f(x_0)\ne0$, <continuity> gives a relative interval of positive length on which $|f(x)|\geq |f(x_0)|/2$. Its <integral> is positive, including when $x_0$ is an endpoint. Thus $\|f\|=0$ implies $f=0$, and this is a <norm> on the given <vector space>.
For the given <sequence>,
$$
\|f_n\|=2\int_0^1x^n\,dx=\frac2{n+1},\qquad \|f_n-f_m\|\leq\frac2{n+1}+\frac2{m+1}.
$$
The last bound tends to zero as both indices tend to infinity, proving the <Cauchy sequence> property. Moreover, \b[the sequence converges to the zero function in this norm]:
$$
\boxed{\|f_n-0\|=\frac2{n+1}\longrightarrow0.}
$$
There is no conflict with the nonzero endpoint values: convergence in this <norm> does not imply <pointwise convergence>.
Back to article page
