Suppose we have $f \in S$, where we identify $S$ as a closed subset of $(C([0,1]), \| \cdot \|_{L^2})$ and $(C([0,1]), \| \cdot \|_{\infty})$. I have been able to show that for some $M > 0$, $$\| f \|_{\infty} \leq M \| f \|_{L^2}.$$
Is it trivially true however that $\| f \|_{L^2} \leq \| f \|_{\infty}$?
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We have $$||f||_\infty = \inf\{l\ge0:|f(x)|\le l \;\mbox{for almost all } x\in[a,b]\}.$$
The 'trivial' bound we can get from this is $$ ||f||_2^2 = \int_a^b|f(x)|^2dx \le ||f||_\infty^2\int_a^bdx = ||f||_\infty^2(b-a)$$
So this only gets you up to a constant in general (for your case with $[a,b]=[0,1]$ it actually gives the true bound.)
The more general case is not as simple.
We know from Jensen's inequality that $||f||_q \le ||f||_p$ for $p\ge q,$ i.e. the norm $||f_p||$ is increasing in $p.$
We can write $$ ||f||_p = \left(\int_a^b|f(x)|^{p-q}|f(x)|^qdx\right)^{1/p}\le ||f||_\infty^{(p-q)/p}||f||_q^{q/p}$$ for any $q<p.$ Taking $p\rightarrow\infty$ gives $$\lim_{p\rightarrow\infty}||f||_p \le||f||_\infty.$$
So $||f||_2 \le ||f||_\infty$ follows from the fact that the $L_p$ norm is increasing in $p$.
Yes it is true. We have: $$ ||f||_{L^2}^2=\int_{[0,1]}|f(x)|^2dx\leq\int_{[0,1]}||f||_{\infty}^2dx=||f||_{\infty}^2 $$ Taking roots gives you the result.