Let $E$ be a vector space and $u \in L(E)$. For the following case, determine whether $u$ has an adjoint with respect to the given bilinear form $\phi (x,y)$.
$E = \{ f: \mathbb{R} \to \mathbb{R}: f $ is a polynomial $\}$,
$\phi(f,g) = \int_0^1 f(t)g(t) dt$
$u(f) = \int_0^1 \frac{f(t)}{\sqrt{t}}dt$
To check if there is an adjoint for u, i have to check for the following:
$(u(f), g) = (f, u'(g))$
So:
$(u(f), g) = (\int_0^1 \frac{f(t)}{\sqrt{t}}dt, g) = \int_0^1 (\int_0^1 \frac{f(t)}{\sqrt{t}}dt)g(t) dt$
and
$(f,u'(g)) = \int_0^1f(t)u'(g)dt$.
Now I will have to check if there exists a $u'$ s.t.
$\int_0^1 (\int_0^1 \frac{f(t)}{\sqrt{t}}dt)g(t) dt$ = $\int_0^1f(t)u'(g)dt$.
How can I do that?
Note that as you define $u(f)$, that is a constant.
Your mistake is that when you compute $\phi(f,g)$, you use the same variable $t$ for two different purposes.
Instead you should have something like
$\int_0^1 u(f)(t)g(t)dt=\int_0^1 [\int_0^1 \frac{f(t')}{\sqrt{t'}} dt']g(t)dt=\int_0^1 \frac{f(t')}{\sqrt{t'}} dt' \cdot\int_0^1g(t)dt$