For $p \in (1, \infty)$, let $\ell^p = \{(x_n) : \sum |x_n|^p < \infty\}$ and let $(\ell^p)^*$ denote the dual space of bounded linear forms on $\ell^p$ $(\mathbb{F} = \mathbb{R})$.
It is straightforward to show that (up to isometric isomorphism), that $\ell^q \subseteq (\ell^p)^*$, but I had more difficulty with proving the reverse inclusion. I thought that it would be easier to prove the contrapositive: $$(y_n) \not \in \ell^q \implies \sup_{||\textbf{x} ||_p \le 1} {\sum |x_n y_n|} = \infty$$
but this defeated me, and the other proofs I have seen since prove the result directly.
Thus, I have two questions. Is there a proof by contradiction of the form described above? (I assume not.)
And more crucially, if I am correct in assuming that there isn't, what are hints that a constructive approach is likely or unlikely to work for problems in analysis?
For instance, it now seems to me that the space of possible $(y_n) \in \mathbb{R}^{\mathbb{N}} \setminus \ell^q$ is probably too vast and irregular to try to construct something suitable for an arbitrary choice.
Nick's idea works, for the first question at least. We let: $$\mathbf{x}^n = (y_1^{q-1}, y_2^{q-1}, \ldots, y_n^{q-1}, 0, 0, \ldots),$$ and define $$\mathbf{u}^n = \frac{\mathbf{x}^n}{\|\mathbf{x}^n\|_p}.$$ We then have $$\langle \mathbf{y}, \mathbf{u}^n\rangle = \frac{1}{\|\mathbf{x}^n\|_p}\sum_{k=1}^n y_k^q.$$ Note that, $$\|\mathbf{x}^n\|_p = \left(\sum_{k=1}^n y^{pq - p}_k\right)^{1/p} = \left(\sum_{k=1}^n y^q_k\right)^{1/p},$$ hence $$\langle \mathbf{y}, \mathbf{u}^n\rangle = \left(\sum_{k=1}^n y^q_k\right)^{1 - 1/p} \to \infty.$$ Thus, $\langle \mathbf{y}, \cdot \rangle$ is not a bounded linear functional on $\ell^p$.