Continuous function and Lp norm
On the vector space $C([0,1])$ of all continuous functions $f : [0,1] \to K$ consider the $p$-norm $$\|f\|_p =\left(\int_0^1 |f(t)|^p dt\right)^{\frac{1}{p}},$$ $f \in C([0,1])$, where $1 \le p < \infty$, as well as the uniform norm $\|f\|_{\infty} = \sup_{t\in[0,1]}|f(t)|$.
I try to show $\|f\|_p \leq \|f\|_\infty$ for all $f \in C([0,1])$ and that $(C([0,1]),\|·\|_p)$ is not complete.
Can please someone help? I am thinking the first part could be releated to the Minkowski inequality.
$\endgroup$ 12 Answers
$\begingroup$Since $|f(x)|\leq \sup_{[0,1]}|f|$ for all $x\in [0,1]$, $$\int_0^1|f(x)|^pdx\leq (\sup_{[0,1]}|f|)^p\int_0^1dx=\|f\|_\infty ^p.$$
Therefore $$\|f\|_{p}=\sqrt[p]{\int_0^1|f(x)|^pdx}\leq \|f\|_{\infty }$$
$\endgroup$ $\begingroup$For the second part, consider for example$$ f_n(x) = \tanh \left(n(x - 1/2)\right).$$As $n \to \infty$, $f_n \to g$ in $L^p$ if $p < \infty$, where$$ g(x) = \begin{cases} & -1 \quad \text{if} \, x \leq 1/2 \\ & 1 \quad \text{if} \, x > 1/2 \end{cases},$$but $g \notin C([0,1])$.
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