$\|f\|_\infty = \sup \{|f(x)| : x \in \mathbb R^n\}$

Question

Let $f$ be a continuous on the measure space $\mathbb R^n,$ $L,$ $\lambda.$ Prove that $\|f\|_\infty = \sup \{|f(x)| : x \in \mathbb R^n\}$

I got "$\leq $" easily, but I have problems using the continuity of $f$ to show it can not be $<$ strictly.

Answer 1

The definition of the $L^{\infty}$ norm I'm assuming you're using is the following:

$$ \|f\|_{\infty}=\inf\{C\geq 0;\,\,|f(x)| \leq C \text{ almost everywhere}\}. $$

Let $\alpha = \sup\{|f(y)|;y\in \mathbb{R}^{n}\} \in [0,\infty]$. Since $|f(x)|\leq \alpha$ for all $x\in\mathbb{R}^{n}$, then $\|f\|_{\infty} \leq \alpha$. Suppose $\|f\|_{\infty}<\alpha$ (assume for the moment that $\alpha<\infty$). Then, there would be a constant $0\leq C<\alpha$ and a full measure set $X \in L$ (meaning that $\lambda(\mathbb{R}^n\setminus X)=0$) such that $$ |f(x)| \leq C $$ for $x\in X$. Since $\alpha$ is the supremum, there is a $y_0\in \mathbb{R}^{n}$ such that $|f(y_0)|>\frac{\alpha+C}{2}>C$ (because $(\alpha+C)/2 < \alpha$). From continuity, of $f$, there is a $\delta>0$ such that if $y\in B_{\delta}(x_0) = \{y\in \mathbb{R}^{n};|y-y_0|<\delta\}$, then $$ |f(y)- f(y_0)|<\frac{\alpha-C}{2}, $$ which means that if $y\in B_{\delta}(y_0)$, $$ |f(y)|>|f(y_0)| - \frac{\alpha-C}{2}>C. $$ Then, $B_{\delta}(y_0) \subset \mathbb{R}^{n}\setminus X$, which contradicts the fact that $\lambda(\mathbb{R}^{n}\setminus X)=0$.

If $\alpha = \infty$ and $\|f\|_{\infty}<\infty$, then the result is simpler, as you can find a point $y_0$ where $|f(y_0)| > C+1$ (say), and use continuity to find a ball where $f$ is greater than $C$ in absolute value, reaching the same contradiction.