As I understand it the norm $\|f\|_2$ on the set of continuous functions on $[0,1]$ is defined by $$\|f\|_2 = \sqrt{\int_0^1|f(t)|^2dt}$$ but what is the infinity norm $\|f\|_\infty$? Is it $$\int_0^1 \max|f(t)|dt$$ so in other words just $\max|f(t)|$?
[Math] the infinity norm on a continuous function space
analysis
Related Solutions
I'll assume we are talking about functions on the real line. Fix $f\in C_0(\mathbb R)$. The fact that $f$ is $C_0$ allows us to write it as a uniform limit of continuous functions with compact support$^1$. So now we can assume without loss of generality that $f$ is continuous with compact support.
Start with $h_0(x)=\begin{cases}e^{-1/x^2},&\ x>0\\0,&\ x\leq0\end{cases}$.
Next you notice that $h(x)=h_0(x)h_0(1-x)\in C^\infty_K$, with support in $[0,1]$. We can normalize it so that $\int_{\mathbb R} h=1$. Then, for each $\varepsilon>0$, you take $$ h_\varepsilon(x)=\frac1\varepsilon\,h\left(\frac x\varepsilon\right) $$ and you note that $\int_{\mathbb R}h_\varepsilon=\int_{\mathbb R}h=1$.
Now form the convolutions $$ f_\varepsilon(x)=\int_{\mathbb R}f(t)\,h_\varepsilon(x-t)\,dt. $$ It is not hard to show that because $h_\varepsilon\in C^\infty$ we have $f_\varepsilon\in C^\infty$; and because $f$ and $h_\varepsilon$ have compact support, so does $f_\varepsilon$.
Finally, given $\varepsilon>0$, as $f$ is continuous with compact support, it is uniformly continuous. So given $c>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<c$ if $|x-y|<\delta$. Then \begin{align*} |f_\varepsilon(x)-f(x)|&=\left|\int_{\mathbb R} [f(x-t)-f(x)]\,h_\varepsilon(t)\,dt\right| \leq\int_{\mathbb R} |f(x-t)-f(x)|\,h_\varepsilon(t)\,dt\\ &\leq c\,\int_{|t|<\delta} h_\varepsilon(t)\,dt +2\|f\|_\infty\,\int_{|t|\geq\delta}h_\varepsilon(t)\,dt\\ &=c +2\|f\|_\infty\,\int_{|t|\geq\delta}h_\varepsilon(t)\,dt. \end{align*} with $\delta$ fixed, the last integral goes to zero as $\varepsilon\to0$ (because the support of $h_\varepsilon$ is contained in $[0,\varepsilon]$). Thus $$ \limsup_{\varepsilon\to0}|f_\varepsilon(x)-f(x)|\leq c $$ for all $x$ and all $c>0$. So $\|f_\varepsilon-f\|_\infty\to0$.
- Fix $f\in C_0(\mathbb R)$. For each $n\in\mathbb N$ there exists $N>0$ with $|f(x)|<\tfrac1n$ when $|x|>N$. Let $g_n$ be continuous, with $0≤g_n≤1$, $g_n(x)=1$ when $|x|≤N$ and $g_n=0$ when $|x|>N+1$. Then $fg_n\in C_c(\mathbb R)$ and $$|f-fg_n|=|f(1-g_n)|<\tfrac1n,$$ so $fg_n\to f$ uniformly.
Best Answer
Up there I think you meant to write
$$ \|f\|_2 := \sqrt{\int_0^1|f(t)|^2dt} $$
The $\sup$-norm is usually defined as follows
$$ \|f\|_\infty := \sup_{x \in [0,1]} |f(x)|$$
Where $f: [0,1] \to \mathbb R$ is continuous, for example. So you get a normed space of functions $(C([0,1]), \|\cdot\|_\infty)$.