We consider the supremum norm $\|f\|=\sup_{x\in [a,b]} |f(x)|$ in the space $B([a,b],\mathbb C)$ of all bounded functions $f: [a,b] \rightarrow \mathbb C$. We obviously have in general that $\|f+g\| \leq \|f\|+\|g\|$ for $f,g \in B([a,b],\mathbb C)$. However, it may happens that for some $f,g$ the equality holds $\|f+g\|=\|f\|+\|g\|$ (for example, but not only then, if $f$ and $g$ are proportional with a positive constant factor).
The problem is: find all $f,g \in B([a,b],\mathbb C)$ such that $\|f+g\|=\|f\|+\|g\|$.
Thanks.
This is essentially the same idea as mathworker21's comment. Equality trivially holds if $f$ or $g$ is 0.
Theorem: If $f$ and $g$ are both not the 0 function, $||f + g|| = ||f|| + ||g||$ if and only if there exists a sequence $x_n\in [a,b]$ such that both
An equivalent characterization is: $||f+g|| = ||f|| + ||g||$ if and only if for all $S\subset [a,b]$ such that $\sup_{x\in S} |f(x)| < ||f||$, we have $\sup_{x\notin S} |g(x)| = ||g||$ and the closure of $\{f(x)/g(x) \mid x\in S\}$ has nonempty intersection with $\mathbb{R}^+$.
Proof: The if direction is straightforward: If 1. and 2. both hold, then note that $||f|| = \lambda ||g||$ for some nonzero $\lambda$ which is a limit point of $f(x_n)/g(x_n)$. Thus there is a subsequence $y_k = x_{n_k}$ such that $\lim f(x_k)/g(y_k) = \lambda$. Furthermore, since the images of $f$ and $g$ are bounded, we can find a subsequence $z_m = y_{k_m}$ such that $\lim f(z_m)$ and $\lim g(z_m)$ both converge. Then: $$\lim |f(z_m)+g(z_m)| = |\lim f(z_m) + g(z_m)| = |(\lambda + 1)\lim g(z_m)| = |1 +\lambda|\cdot ||g|| = ||g|| + \lambda ||g|| = ||f|| +||g||$$ so $\sup |f+g| \ge ||f|| +||g||$, which implies equality.
For the only if direction, first suppose there is no sequence satisfying 1. Then there must exist $S\subset [a,b]$ be a set such that $\sup_{x\notin S} |f(x)| < ||f(x)||$ and $\sup_{x\in S} |g(x)| < ||g||$. We can construct such an $S$ by taking decreasing sequences of sets $A_n = \{x : |f(x)| > ||f||-\frac1n\}$ and $B_n = \{x : |g(x)| > ||g||-\frac1n\}$. For sufficiently large $n$, these are disjoint, because otherwise one could pick a sequence $x_n\in A_n\cap B_n$ and it would satisfy 1. Then for $n$ large enough that $A_n \cap B_n = \emptyset$, we can simply take $S=A_n$. Once we have such an $S$, clearly $\sup_{x\in S} |f(x) + g(x)| < ||f|| + ||g||$ and $\sup_{x\notin S} |f(x) + g(x)| < ||f|| + ||g||$, so of course $\sup_{x\in[a,b]} |f(x)+g(x)| < ||f||+||g||$.
To complete our proof, suppose there exists a sequence such that 1. holds but not 2. Let $A_n$ and $B_n$ be as defined above. Let $u_n = \sup_{x\in A_n\cap B_n} \cos(\arg(f(x)/g(x)))$. By assumption, $\lim u_n < 1$ (otherwise the ratio $f(x)/g(x)$ would get arbitrarily close to lying in $\mathbb{R}^+$ at the same time as $f$ and $g$ get close to their maximal absolute values, so we could find a sequence such that 2. holds). If for some set $S\subset [a,b]$, $\sup_{x\in S} \cos(\arg(f(x)/g(x))) <1$, then $\sup_{x\in S} |f(x) + g(x)| < ||f|| + ||g||$. This is intuitively clear, to prove it formally let $\alpha\in(0,1)$ be such that $\sup_{x\in S}\cos(\arg(f(x)/g(x))) \le \alpha$. Then \begin{eqnarray} \sup_{x\in S} |f(x) + g(x)| &=& \sup_{x\in S}\sqrt{|g(x)|^2 + |f(x)|^2 + 2 |f(x)g(x)| \cos\left(\arg\left(\frac{f(x)}{g(x)}\right)\right) }\\ &\le&\sup_{x\in S}\sqrt{|g(x)|^2 + |f(x)|^2 + 2 |f(x)g(x)| \sup_{x\in S}\left(\cos\left(\arg\left(\frac{f(x)}{g(x)}\right)\right)\right) }\\ &\le&\sup_{x\in S}\sqrt{|g(x)|^2 + |f(x)|^2 + 2 |f(x)g(x)| \alpha }\\ &=& \sup_{x\in S} \sqrt{(|g(x)| + |f(x)|)^2 - (1-\alpha)2|f(x)g(x)|}\\ &\le& \sup_{x\in S} |g(x)| + |f(x)| - \frac{2(1-\alpha) |f(x)g(x)|}{|f(x)| + |g(x)|} \\ &\le& ||g|| + ||f|| - \frac{2(1-\alpha)||f||\space||g||}{||f|| + ||g||} < ||f||+||g|| \end{eqnarray} (the square root is expanded in the 2nd last line using the fact that $\sqrt{a-b} < \sqrt{a} - \frac{b}{2\sqrt{a}}$ for $0<b<a$). Since $\alpha$ is defined by being an upper bound, we can assume it is close enough to 1 so that this step works.
This completes the proof, because for $n$ sufficiently large, we must have $u_n< 1$ (and this is a decreasing sequence), so $\sup_{x\in A_n\cap B_n} |f(x) + g(x)| < ||f|| + ||g||$. By the definition of $A_n$ and $B_n$, we also have $\sup_{x\notin A_n\cap B_n} |f(x) + g(x)| < ||f|| + ||g||$ so the triangle inequality is strict if 1. holds but not 2. ∎
Addendum after Alex's comment:
Corollary: $||f+g|| = ||f|| + ||g||$ if and only if there exists a sequence $z_n$ such that $$ \lim \log\left(f(z_n) \overline{g(z_n)}\right) = \log ||f|| + \log ||g||\tag{∗} $$ Note, if $f$ or $g$ is 0, then (∗) holds because both sides of the equation are $-\infty$. Since $f$ and $g$ are bounded, there will never be the issue of $\infty-\infty$.
Proof: We show that (∗) is equivalent to hypotheses 1. and 2. of the theorem. We can see right away that (∗) implies 1. since $$\Re\left(\log\left(f(z_n) \overline{g(z_n)}\right)\right) = \log|f(z_n) \overline{g(z_n)}| = \log |f(z_n)| + \log |g(z_n)|$$ if $\lim |f(z_n)| < ||f||$ or $\lim |g(z_n)| < ||g||$, then we would have $\lim \Re\left(\log\left(f(z_n) \overline{g(z_n)}\right)\right) < \log||f|| + \log||g||$. The imaginary part of the logarithm shows that (∗) also implies 2., because $$\Im\left(\log\left(f(z_n) \overline{g(z_n)}\right)\right) = \arg\left(f(z_n) \overline{g(z_n)}\right) = \arg\left(\frac{f(z_n)}{g(z_n)}\right)$$ so if the limit of the imaginary part is 0, then the limit of $f(z_n)/g(z_n)$ has argument 0 and hence is in $\mathbb{R}^+$. To see that 1. and 2. together imply (∗), we note that since $f(x_n)/g(x_n)$ has a limit point in $\mathbb{R}^+$, there must exist a subsequence $z_k = x_{n_k}$ such that $\lim f(z_k)/g(z_k) = L\in\mathbb{R}^+$. Then this $z_k$ satisfies (∗):\begin{eqnarray} \lim \log\left(f(z_k)\overline{g(z_k)}\right) &=& \lim \log |f(z_k)\overline{g(z_k)}| + i\lim \arg\left(f(z_k)\overline{g(z_k)}\right)\\ &=& \lim \log|f(z_k)| + \lim\log |g(z_k)| + i \lim \arg\left(\frac{f(z_k)}{g(z_k)}\right)\\ &=& \log||f|| + \log||g|| + i\arg(L)\\ &=& \log||f|| + \log ||g|| \end{eqnarray}