Showing an inequality with specific norms is false

Question

Consider the linear space $C([a,b])$ for a compact interval $[a,b] \subset \mathbb{R}$.

For $f \in C([a,b])$, let

$$ \lvert \lvert f \rvert \rvert_{1} = \int_{a}^{b} \lvert f \rvert $$

$$ \lvert \lvert f \rvert \rvert_{\infty} = \underset{x \in [a,b]}{\text{max}} \lvert f(x) \rvert$$

Then $\lvert \lvert \cdot \rvert \rvert_{1}$ and $\lvert \lvert \cdot \rvert \rvert_{\infty}$ each define norms on $C([a,b])$.

Show that there is no constant $c \geq 0$ such that $\lvert \lvert f \rvert \rvert_{\infty} \leq c \lvert \lvert f \rvert \rvert_{1}$ for all $f \in C([a,b])$.

My approach is to fix a $ c \geq 0$, and then find a continuous function on $[a,b]$, that may be dependent on $c$, such that

$$ \underset{x \in [a,b]}{\text{max}} \lvert f(x) \rvert > c * \int_{a}^{b} \lvert f \rvert $$

but I am struggling with this. Any help would be appreciated, thank you!

Answer 1

An idea: suppose the above inequality between $\| \cdot \|_{1} $ and $\| \cdot \|_{\infty}$ does hold. Then a convergent sequence in $(C^0([a,b]),\| \cdot \|_{1})$ would also converge in $(C^0([a,b]),\| \cdot \|_{\infty})$. Note that from this exercise it follows that $\| \cdot \|_{1} $ and $\| \cdot \|_{\infty}$ are not equivalent norms on $(C^0([a,b])$.

Answer 2

Let's consider the interval $[-1,1]$ for simplicity. The argument for any $[a,b]$ is the same.

Let $f_n$ be a function which graph is a broken line joining $(-1,0)$, $\left(-\frac{1}{n},0\right)$, $(0,1)$, $\left(\frac{1}{n},0\right)$ and $(1,0)$. Then $\|f_n\|_1=\frac{1}{n}\to 0$, while $\|f_n\|_{\infty}=1$. Thus $(f_n)$ tends to $0$ function in the norm $\|\cdot\|_1$, while $(f_n)$ does not tend to $0$ in the maximum norm. The existence of a $c$ you are asking for would imply that a sequence tending to $0$ in the integral norm also tends to zero in the maximum norm.