Showing $\|f \|_2$ isn't equivalent to $\| f\|_\infty$ Was just looking for some help with this question, we fix $a<b$ in $\mathbb R$ and consider the two norms on the vector space $C[a,b]$ of continuous functions , in this instance I am defining $$\| f \|_2 = \sqrt{ \int_a^b |f|^2}$$
$$\| f \|_\infty = \sup \{ |f(x)| : a \leq x \leq b \} $$
and I am struggling to find a function to show that there is no $K$ such that $$\| f \|_\infty \leq K\| f \|_2$$ 
any help greatly appreciated thanks
 A: Take $a=0,b=1$ for convenience. Let $f_n(x) = x^n.$ Then $\|f_n\|_\infty = 1$ for all $n.$ But
$$\|f_n\|_2 = (\int_0^1x^{2n}\, dx)^{1/2} = \left (\frac{1}{2n+1}\right )^{1/2}.$$
Thus $\|f_n\|_2\to 0,$ so an inequality of the form $1=\|f_n\|_\infty\le K \|f_n\|_2$ for all $n$ is impossible.
A: It's easiest to describe a set of appropriate functions geometrically, rather than write down a formula. Take a function such that its square has a graph that looks like an isoceles triangle of width $2/n$ and height $n$, with the base on the $x$-axis. The $L^2$-norm of this function is the square root of the area of the triangle, i.e. $1$. Taking the square root, the supremum is $\sqrt{n}$. Hence if there were such a $K$, you would have
$$ \sqrt{n} \leqslant K \cdot 1, $$
which is obviously false as we can take $n$ as large as we like.
A: First, you will have no luck finding one single function $f \in C[a,b]$ such that there is no $K \geq 0$ with $\|f\|_{\infty} \leq K \|f\|_2$. The reason is that for $f \in C[a,b]$, it holds that $\|f\|_2 = 0 \iff f \equiv 0 \iff \|f\|_{\infty} = 0$. (This is not true if the search was expanded to measurable functions, rather than restricting to continuous functions, since you could take $f$ as the indicator on the rationals in $[a,b]$). But then, if $\|f\|_2 \neq 0$, you can just defined $K = \frac{\|f\|_{\infty}}{\|f\|_2}$. That said, as suggested in the comments above, you can find a sequence of functions $f_n \in C[a,b]$ such that $\inf\{K \geq 0 \,\mid\, \|f_n\|_{\infty} \leq K \|f_n\|_2\} = \infty$. For example, if $[a,b] = [0,1]$ try
$$
f_n(x) = 
\begin{cases}
-2^n \,x + 1 & x \in [0,2^{-n}]\\
0 & x \in (2^{-n}, 1]
\end{cases}
$$
(adjust accordingly for $x \in [a,b]$). 
