Consider the space $C([a,b])$ of all continuous functions $f\colon [a,b]\rightarrow \mathbb{R}.$ Show that the function $\|\cdot\|_p\colon C([a,b]) \rightarrow [0,\infty),p>1$, given by $$ {\|f\|}_p = \sqrt[p]{\int_{a}^{b} |f(x)|^p {\rm d}x}\quad (f\in C([a,b]))$$ is a norm.
Apparently we have to prove triangle inequality...(?) (Note:hits said one can use Holder. Is there a simpler way?)
Guys please help!
Thank you!
The inequality which you want to show is often called the Minkowski inequality. An unusual proof can be found here Link.
Below I will show a standard one which involves Hölder's inequality.
We want to show the triangle inequality, that is, that for any $f$, $g \in C([a, b])$
$$\begin{align*}\| f+g\|_p = \left(\int_{a}^b |f(x)+g(x)|^{p} \ \mathrm{d}x\right)^{\frac{1}{p}} &\leq \|f\|_p + \|g\|_p \\ &= \left(\int_{a}^b |f(x)|^{p} \ \mathrm{d}x\right)^{\frac{1}{p}}+\left(\int_{a}^b |g(x)|^{p} \ \mathrm{d}x\right)^{\frac{1}{p}}\end{align*}.$$
Let $p>1$ and let $q$ be such that $\frac{1}{p}+ \frac{1}{q}=1$.
We use Hölder's inequality, that is, if $p$, $q > 1$ and $\frac{1}{p}+ \frac{1}{q}=1$, then $$ \int_a^b |f(x)g(x)| \ \mathrm{d}x \leq \left(\int_a^b |f(x)|^p \ \mathrm{d}x \right)^{\frac{1}{p}} \left(\int_a^b |g(x)|^q \ \mathrm{d}x \right)^{\frac{1}{q}},$$ where $f$ is $p$-integrable and $g$ is $q$-integrable (which holds if $f, g \in C([a, b])$).
First note that $$|f+g|^p = |f+g||f+g|^{p-1} \leq (|f|+|g|)|f+g|^{p-1}= |f||f+g|^{p-1}+|g||f+g|^{p-1} $$ (since triangle inequality holds for absolute value $| \cdot |$).
Now by applying the above we obtain $$ \begin{align*} \int_a^b |f(x) + g(x)| \ \mathrm{d} x \leq \int_a^b |f(x)||f(x)+g(x)|^{p-1} \ \mathrm{d}x + \int_a^b |g(x)||f(x)+g(x)|^{p-1} \ \mathrm{d}x\end{align*}.$$
Further, we make a use of Hölder's inequality and an easy observation that $(p-1)q = p$. Therefore,
$$ \begin{align*} LHS=\int_a^b |f(x) + g(x)| \ \mathrm{d} x &\leq \int_a^b |f(x)||f(x)+g(x)|^{p-1} \ \mathrm{d}x + \int_a^b |g(x)||f(x)+g(x)|^{p-1} \ \mathrm{d}x \\ & \leq \left(\int_a^b |f(x)|^p \ \mathrm{d}x \right)^{\frac{1}{p}}\left(\int_a^b |f(x)+g(x)|^{(p-1)q} \ \mathrm{d}x \right)^{\frac{1}{q}} \\ &+\left(\int_a^b |g(x)|^p \ \mathrm{d}x \right)^{\frac{1}{p}}\left(\int_a^b |f(x)+g(x)|^{(p-1)q} \ \mathrm{d}x \right)^{\frac{1}{q}} \\& = \left(\int_a^b |f(x)|^p \ \mathrm{d}x \right)^{\frac{1}{p}}\left(\int_a^b |f(x)+g(x)|^{p} \ \mathrm{d}x \right)^{\frac{1}{q}} \\ &+\left(\int_a^b |g(x)|^p \ \mathrm{d}x \right)^{\frac{1}{p}}\left(\int_a^b |f(x)+g(x)|^{p} \ \mathrm{d}x \right)^{\frac{1}{q}} \\ &= \left(\left(\int_{a}^b |f(x)|^{p} \ \mathrm{d}x\right)^{\frac{1}{p}}+\left(\int_{a}^b |g(x)|^{p} \ \mathrm{d}x\right)^{\frac{1}{p}}\right)\left(\int_a^b |f(x)+g(x)|^{p} \ \mathrm{d}x \right)^{1- \frac{1}{p}}=RHS \end{align*}.$$
To finish we devide LHS and RHS by $\left(\int_a^b |f(x)+g(x)|^{p} \ \mathrm{d}x \right)^{1- \frac{1}{p}}$.