Let $f \in L^2([0,1])$.
Assume that for every $\varphi \in W^{1,2}([0,1])$ we know that $$ \int_{0}^1 \varphi'(x) f(x) dx \leq \|\varphi \|_{W^{1,2}}. $$ Does this imply that $f$ belongs to $W^{1,2}([0,1])$?
My guess is yes because we can define $f'$ as $$ \int_0^1 \varphi(x) f'(x) dx = - \int_0^1 \varphi'(x) f(x) dx, $$ but I am not able to formalize this idea. Any hint is welcome, thank you
On
In order to define a function $g \in L^2$ via the formula $$\int_0^1 \phi \, g \, \mathrm d x = \int_0^1 \phi' \, f \, \mathrm{d}x$$ you need $$\int_0^1 \phi' \, f \, \mathrm{d} x \le C \, \|\phi\|_{L^2}.$$ In this case, the right-hand side of the first formula is defined on $W^{1,2}$ and is continuous w.r.t. the $L^2$-norm. Since $W^{1,2}$ is dense in $L^2$, it extends uniquely to a continuous functional on $L^2$. By the Riesz representation theorem, this can be identified with an $L^2$-function $g$.
The answer is no. Consider the following function $f: [0,1] \to \Bbb R$: $$f(x) = \begin{cases} 0 & x < 1/2 \\ 1 &x \ge 1/2 \end{cases} $$ You can easily check that $f \in L^2$.
Moreover, for all $\varphi$ $$\int_0^1 \varphi'(x) f(x) \ \mathrm dx \le \int_0^1 | \varphi'(x) | \cdot 1 \ \mathrm dx \le ||\varphi||_{W^{1,2}}$$ However $f$ is a step function: it has no weak derivative.