*Apologies if this post is shoddily written; I mainly just wanted to record an explicit counterexample before I forget it, and this post is really just meant for future me and not for general consumption.*

A common construction in interpolation theory involves starting with two Banach spaces $V$ and $W$, and building the space $U = V+W$ with norm \begin{equation} \|u|_{U} := \inf_{u = v+w} \|v\|_{V} + \|w\|_{W}. \end{equation} For this to work, it is crucial that $V$ and $W$ can both be continuously embedded into some larger Hausdorff topological vector space.

So what happens when we don't make this assumption?

The answer is that the resulting "norm" make actually fail to be a norm. Recall that a norm needs to be non-negative, non-degenerate, absolutely homogeneous, and satisfying triangle inequality. Non-negativity of $\|u\|_{U}$ is obvious, as is its absolute homogeneity. The fact that triangle inequality is satisfied comes from the following computation. Given $u_1, u_2\in U$, for any pair of decompositions $u_1 = v_1 + w_1$ and $u_2 = v_2 + w_2$, by definition we have \[ \|u_1 + u_2\|_U \leq \|v_1 + v_2\|_{V} + \|w_1 + w_2\|_{W} \leq \|v_1\|_{V} + \|v_2\|_V + \|w_1\|_W + \|w_2\|_W.\] Taking the infimum on the right hand side yields triangle inequality.

However, what about non-degeneracy? We argue by contradiction: suppose a non-zero element $u$ has $U$-norm zero, then there must exist a sequence of elements $v_i$ and $w_i$ of $V$ and $W$ respectively, such that $v_i + w_i = u$ and $v_i\to 0$ and $w_i\to 0$ in their respective norms. However, since $V$ and $W$ are both continuously embedded into some larger TVS, this means that $v_i$ and $w_i$ both converge to $0$ in the larger TVS, and hence their sum also converge to $0$. But noting that $v_i + w_i$ is a constant sequence, this means that $u$ must be the zero element in the larger TVS, which gives a contradiction.

Note that in this argument we needed to assume the existence of a larger TVS. So non-degeneracy is the one condition that can break if we don't make this assumption.

To illustrate that this is a real problem, I'll provide below an example where the non-degeneracy fails. Note that the arguments above never actually required completeness of $V$ and $W$, so we'll just work with normed linear spaces and not their completions for convenience.

In our case, we will take the underlying vector spaces $V$ and $W$ to be equal; then the set $U$ is also equal to them. Let $X = \mathbb{N}\cup \{\infty\}$. Consider first the set $\mathbb{R}^X$. The restriction to $\mathbb{N}$ may be regarded as a real valued sequence. We will let $V$ be the subspace satisfying the conditions $f\in V$ if and only if

- $f(1) = 0$
- and the sequence $f|_{\mathbb{N}}$ converges

Note that these conditions are closed under linear combination and hence this makes a vector subspace.

We equip this space with two norms. The first norm is \[ \|f\|_{1} := \sup \left( \{ |f(i) - f(i+1)| : i\in \mathbb{N} \} \cup \{ | (\lim f|_{\mathbb{N}}) - f(\infty) | \} \right) \] Note that as $f|_{\mathbb{N}}$ is a convergent sequence, this norm is finite on $V$. And it is a norm as $\|f\|_{1} = 0 \implies f$ is constant. The second norm is \[ \|f\|_{2} := \sup \left( \{ 2^{-i} |f(i)| : i \in \mathbb{N} \} \cup \{ |f(\infty)| \} \right). \]

Now let $f$ be the function that is identically 0 except $f(\infty) = 1$. I claim that \[ \inf_{f = g+h} \|g\|_1 + \|h\|_2 = 0. \] Let \[ g_k(i) = \begin{cases} 0 & i \leq k \newline i/k - 1 & k < i \leq 2k \newline 1 & i > 2k \end{cases} \] with $g_k(\infty) = 1$, and set $h_k = f - g_k$ (so $h_k(\infty) = 0$). Observe that \[ \|g_k\|_1 = 1/k, \quad \|h_k\|_2 \leq 2^{-k}. \] This shows that were we to define a sum "norm" using the formula given, this would be at best a seminorm and not a true norm.

Lastly, let me just indicate why the two norms are not compatible in the sense of allowing to be embedded in a larger TVS. Basically, if both norms embed, then if a sequence is convergent with respect to both norms, the two limits must be the same by the uniqueness of limits in the larger Hausdorff TVS. But the sequence of functions $g_k$ above converges in the first norm to $0$, while it converges in the second norm to $f$.