Projection onto space in $L^2()$ gives shorter length

Let $f_1,f_2,\ldots,f_n\in L^2([0,1])$, and let $V$ denote their span. Let $P:L^2([0,1])\rightarrow V$ be the projection onto $V$.

Let $g\in L^2([0,1])$. Suppose also that $g\in L^p([0,1])$ for some $1\leq p<\infty$. Is it always true that $\|Pg\|_p\leq\|g\|_p$?

Solutions Collecting From Web of "Projection onto space in $L^2()$ gives shorter length"

The answer is no.

That seems to be a overkill, but we can use the fact that $||f||_p \to ||f||_\infty$ as $p\to \infty$ and $||f||_p \leq ||f||_q$ if $p \leq q$. If we can find $g\in L^2(0,1)$ such that $||Pg||_\infty > ||g||_\infty$, then for $p$ big enough, we have $||Pg||_p > ||g||_p$.

I tried $g = \chi_{[0,1/2]}$, $h= g+ 2\chi_{[23/24, 1]}$ and project $g$ onto $h$:

$$Pg= \frac{\langle g, h\rangle}{||h||^2_2}h$$

and seems we have $||Pg||_\infty = 3/2> ||g||_\infty.$