Non-existence of weak solution in one dimension Let $\Omega=(1,\infty).$ Then for any given $f\in L^2(\Omega),$ the equation 
$$
-u''=f\,\,\text{in}\,\,\Omega,
$$
does not admit any weak solution in $W_{0}^{1,2}(\Omega).$
I tried the solution by contradictory argument. Indeed, suppose such a weak solution exists, say $u.$ Then for every $\phi\in H_0^{1}(\Omega),$ we have
$$
\int_{\Omega}\,u'\phi'\,dx=\int_{\Omega}f\phi\,dx
$$
I tried to construct $\phi_n\in C_c^{\infty}(\Omega)$ such that the above inequality become false. I think this idea will work, but I am still unable to construct such $\phi.$ If anyone can help me regarding this or with some other idea on solving this problem, it will be very grateful for me. Thanks.
 A: Let me take $\Omega=(0,+\infty)$ for convenience. Define 
$$
u(x) = \min(x,x^{-1/2}).
$$
Then $u\not\in L^2(\Omega)$ and $u\in L^1_{loc}(\Omega)$. It has first and second weak derivatives $u'$ and $u''$ given by
$$
u'(x) = \chi_{(0,1)} -\frac12 \chi_{(1,\infty)} x^{-3/2}
$$
and
$$
u''(x) = \frac34 \chi_{(1,\infty)} x^{-5/2}.
$$
It follows $u',u''\in L^2(\Omega)$. In addition, $u$ (rather $u'$) satisfies the weak formulation. Moreover, the weak formulation has at most one solution since $\int_\Omega (u')^2=0$ implies $u=const$ hence $u=0$.
Now take $f=-u''$. Then you have $f\in L^2(\Omega)$ with non-existent weak solution in $H^1(\Omega)$. 
A: I tried to sow some doubt about the question yesterday, by giving a couple of examples, where a weak solution actually exists in the comments.
Below is a failed attempt to show that a weak solution always exists, but as pointed out by daw in a comment I failed at properly checking the coercivity. You can consider it as an example on how not to prove something.
For every $f\in L^2(\Omega)$ the equation
$$
-u'' = f
$$
admits a weak solution $u\in H^1_0(\Omega)$.
By multiplying with test functions and partial integration you get the weak formulation:
\begin{equation}\tag{1}
a(u,v) := \langle u',v'\rangle_{L^2} = \langle f, v\rangle_{L^2},
\end{equation}
for all $v \in C^\infty_0(\Omega)$. By density this statement also holds for all $v \in H^1_0(\Omega)$. One easily verifies that $a$ is a coercive (this is not true), bounded sesquilinear form on $H^1_0(\Omega)$. The theorem of Lax-Milgram shows that $u\in H^1_0(\Omega)$ such that (1) holds exists if $\langle f, \cdot \rangle_{L^2} \in (H^1_0(\Omega))'$. That $\langle f, \cdot \rangle_{L^2} \in (H^1_0(\Omega))'$ is clear by the Cauchy-Schwarz inequality.
