I have shown that $\lVert f\rVert_2\leq \sqrt{b-a}\cdot \lVert f\rVert_\infty$, $\forall f\in C[a,b]$.
Now I want to show that there is no positive constant $m$ such that $m\|f\|_{\infty}\leq \|f\|_2$, $\forall f\in C[0,1]$.
It is given also the hint to consider the functions $$f_k(x)=\begin{cases}1-kx, & x\in \left [0, \frac{1}{k}\right ] \\ 0 , & x\in \left [\frac{1}{k}, 1\right ]\end{cases}$$ for $k=1, 2, \ldots$
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Could you explain to me how we could get teh desired result considering tehse functions?
Hint: $$ \Vert f_k \Vert_\infty = 1, \quad \text{for all } k, $$ whereas $$ \Vert f_k \Vert_2 = \left(\int_0^{1/k} (1-kx)^2\right)^{1/2} \to 0,\quad \text{as } k\to\infty. $$ Now assume that there exists an $m > 0$ such that $$ m\Vert f_k \Vert_\infty \leq \Vert f_k \Vert_2, $$ and let $k\to\infty$ on both sides to reach a contradiction.