# The limit of p-norms is the max norm in the space of continuous functions in [a,b]

1. Jan 20, 2012

### ELESSAR TELKONT

1. The problem statement, all variables and given/known data

$\lim_{p\rightarrow\infty}\left(\int_{a}^{b}\vert x(t)\vert^{p}\,dt\right)^{\frac{1}{p}}=\Vert x\Vert_{\infty}$

2. Relevant equations

\begin{align*} \Vert x\Vert_{s}\leq (b-a)^{\frac{r-s}{rs}}\Vert x\Vert_{r}\,\forall 1\leq s<r<\infty\\ \Vert x\Vert_{s}\leq (b-a)^{\frac{1}{s}}\Vert x\Vert_{\infty}\,\forall 1\leq s<\infty\\ 0\leq\frac{\vert x(t)\vert}{\Vert x\Vert_{\infty}}\leq 1\,\forall t\in [a,b] \end{align*}

3. The attempt at a solution

Using the last of the former inequalities I can say that

\begin{align*} \left(\int_{a}^{b}\frac{\vert x(t)\vert^{p}}{\Vert x\Vert_{\infty}^{p}}\,dt\right)^{\frac{1}{p}}\leq (b-a)^{\frac{1}{p}} \end{align*}

and since in the limit
\begin{align*} \lim_{p\rightarrow\infty}a^{\frac{1}{p}}=1\, \forall a>0 \end{align*}

then

\begin{align*} \lim_{p\rightarrow\infty}\left(\int_{a}^{b}\frac{ \vert x(t)\vert^{p}}{\Vert x\Vert_{\infty}^{p}}\,dt\right)^{\frac{1}{p}}\leq 1 \end{align*}

But since I have the other relations between the norms and these relations say that, fixed a function $x(t)$, the norms as a function or $p$ are a monotonic function that grows, then the last inequality say that in the limit the integral equals to one. Therefore, since I can write

\begin{align*} \Vert x\Vert_{p}=\Vert x\Vert_{\infty}\left(\int_{a}^{b}\frac{\vert x(t)\vert^{p}}{\Vert x\Vert_{\infty}^{p}}\,dt\right)^{\frac{1}{p}} \end{align*}

it follows the proposition taking the limit by the former reasoning.

MY PROBLEM IS IF IT IS TRUE MY ASSUMPTION OF THE MONOTONICITY OF THE NORMS WITH P. IS IT VALID THE ARGUMENT? IF NOT, WHAT OTHER ARGUMENT MUST BE USED? THANKS IN ADVANCE.