What a proof of weak convergence is supposed to look like Recall the definition of weak convergence: if $u_{k}\in \mathcal{B}$, we say that $u_{k}$ converges weakly to $u$ if $\forall \ell \in $ the dual space $\mathcal{B}^{*}$ ($\mathcal{B}$ is a Banach space) implies that $\langle \ell, u_{k}\rangle \to \langle \ell, u\rangle$.
Now, for $u$ any fixed function in $C^{\infty}_{0}(\mathbb{R})$ which is not identically 0, I need to show that the sequence of functions $u_{k}(x)=u(x+k)$ converges weakly to $0$ in $L^{2}(\mathbb{R})$, but which does not converge in the usual sense.
The only problem is, I have absolutely no idea what this is supposed to look like - the only other problem I've ever seem like this was one where $u_{k}$ was an orthonormal basis for the space it was in, so that was ridiculously easy to show. 
For that one, it basically involved something like showing that the sum of the inner product converged, and do that it's general term went to zero, but I'm not sure what that would look like in this case. If somebody would be willing to do this out in a detailed manner, just so I could see how a problem like this is supposed to be solved, I would appreciate it so very much!
 A: There is some $M$ such that if $|x| > M$, then $u(x) = 0$. Also, there is some $B$ such that $|u(x)| \le B$ for all $x$. Hence
$|u(x)| \le  B \cdot 1_{[-M,M]}(x)$, and so
$|u_k(x)| \le  B \cdot 1_{[-M,M]}(x+k) =  B \cdot 1_{[-M-k,M-k]}(x)$.
The trick is to find a suitable representation for a member of the dual space, and this is furnished by the Riesz representation theorem.
If $\phi \in L^2(\mathbb{R})^*$, then we can write $\phi(x) = \langle f, x \rangle$, for some $f \in L^2(\mathbb{R})$, and using the above  we get
\begin{eqnarray}
|\phi(u_k)| &=& | \langle f, u_k \rangle | \\
& \le & \int |f(x)|  |u_k(x)| dx \\ 
& = & \int |f(x)| 1_{[-M-k,M-k]}(x) |u_k(x)| dx \\ 
&\le& \|f \cdot 1_{[-M-k,M-k]} \| \|u_k\| \\
&=& \sqrt{\int_{ [-M-k,M-k]  } |f|^2 } (2MB)
\end{eqnarray}
Since $f \in L^2(\mathbb{R})$, we see that $\lim_{k \to \infty} \sqrt{\int_{ [-M-k,M-k]  } |f|^2 } = 0$, and so $u_k$ converges weakly to $0$.
Since $\|u_k\| = \|u\|$ for all $k$, then if $u \neq 0$, we see that the $u_k$ do not converge (if they did, then they would also converge weakly to the same limit).
A: If $u\in L^2(\mathbb R)$, then for all $\varepsilon>0$, there exists an $n=n(\varepsilon)>0$, such that
$$
\int_{|x|>n}|u|^2\,dx<\varepsilon^2.
$$ 
Why? Because, the function
$$
u_n(x)=\left\{\begin{array}{lll}
u(x) &\ \text{if} & |x|\le n, \\
0 &\ \text{if} & |x|> n,
\end{array}\right.
$$
converges pointwise to $u$, and $\int_{\mathbb R}|u_n|^2\le\int_{\mathbb R}|u|^2$, and using
Lebesgue Dominated Convergence Theorem we obtain that 
$$
\int_{|x|> n}|u(x)|^2\,dx=\|u_n-u\|^2<\varepsilon^2,
$$
and hence $\lim_{n\to\infty}\|u_n-u\|=0$.
So, if $\varphi\in C_0^\infty(\mathbb R)$, and supp$\,\varphi\subset [-N,N]$, and $k>N+n(\varepsilon)$, then
\begin{align}
(u_k,\varphi)&=\int_{-\infty}^\infty u_k(x)\,\varphi(x)\,dx=\int_{-N}^N u(x+k)\,\varphi(x)\,dx=\int_{-N+k}^{N+k} u(x)\,\varphi(x-k)\,dx,
\end{align}
and hence
$$
|(u_k,\varphi)|\le \|\varphi\|\,\left(\int_{-N+k}^{N+k}u^2(x)\,dx\right)^{1/2}\le
\|\varphi\|\,\left(\int_{x>n(\varepsilon)}u^2(x)\,dx\right)^{1/2}\le
\varepsilon\,\|\varphi\|.
$$
This means that, $\lim_{k\to\infty}(u_k,\varphi)=0$, and thus $u_k\to u$ weakly. Not strongly though, as $\|u_k\|\not\to 0$.
