Suppose that $C$ is a convex subset of a normed vector space $X$. If $f:C\to \mathbb{R}$ is continuous and there exists a $\beta\in \left(0,1\right)$ for which $$ f(\beta x+(1-\beta)y)\leq \beta f(x)+(1-\beta)f(y) $$ for all $x,y\in C$. Is it true that $f$ is convex?
I tried following the classical proof that midpoint convexity and continuity imply convexity, but so far I failed.
Judging by the exchange in the comments, it seems that you just need to show that $$f(\alpha x + (1 - \alpha)y) \le \alpha f(x) + (1 - \alpha) f(y) \tag{$\star$}$$ holds for $\alpha$ on a dense subset of $[0, 1]$. Let's define this set by saying: \begin{align*} P_0 &= \{0, 1\} \\ P_{n+1} &= P_n \cup \{\beta c + (1- \beta) d : c,d \in P_n\}, \quad \forall n \ge 0 \\ P &= \bigcup_{n=0}^\infty P_n. \end{align*} We can prove by induction that, if $\alpha \in P_n$, then $(\star)$ holds. It would then follow that $(\star)$ holds for $\alpha \in P$, which we would need to prove is dense. Note that $(\star)$ holds for $\alpha \in P_0$ trivially.
Suppose $n \ge 0$ is such that $\alpha \in P_n$ implies $(\star)$ is true. If $\alpha \in P_{n+1}$, then either $\alpha \in P_n$ (in which case $(\star)$ holds true by the inductive hypothesis), or $$\alpha = \beta c + (1 - \beta)d$$ for some $c, d \in P_n$. Using the one-weight convexity condition, \begin{align*} &f(\beta(cx + (1 - c)y) + (1 - \beta)(dx + (1 - d)y)) \\ \le\;&\beta f(cx + (1 - c)y) + (1 - \beta)f(dx + (1 - d)y) \\ \le\;&\beta(cf(x) + (1 - c)f(y)) + (1 - \beta)(df(x) + (1 - d)f(y)) \tag{$\dagger$} \\ =\;&(\beta c + (1 - \beta)d)f(x) + (1 - \beta c - (1 - \beta)d)f(y), \end{align*} where $(\dagger)$ is true by the induction hypothesis. This shows, by induction, that if $\alpha \in P_n$ for some $n$ (i.e. if $\alpha \in P$), then $(\star)$ holds.
The construction of $P$ gives us the following important property: $$c, d \in P \implies \beta c + (1 - \beta)d \in P.$$ This holds true because $c$ and $d$ must lie in $P_n$ and $P_m$ respectively, for some $m, n \ge 0$. But, we constructed these $P_i$ sets to be nested, so both must lie in $P_{\max\{m,n\}} \subseteq P$. We now just need to show $P$ is dense in $[0, 1]$.
Suppose, for the sake of contradiction, that $P$ is not dense in $[0, 1]$. Then there is some non-empty open subinterval $(c, d)$ of $[0, 1]$ that contains no points of $P$. Let \begin{align*} c' &= \inf\{a \in [0, c] : (a, d) \cap P = \emptyset\} \\ d' &= \sup\{a \in [d, 1] : (c, a) \cap P = \emptyset\}. \end{align*} I claim that $(c', d') \cap P = \emptyset$. If we had $b \in (c', d') \cap P$, then $c' < b < d'$. As $b \in P$, we know $b \notin (c, d)$, so either $c' < b \le c$ or $d \le b < d'$.
Suppose it's the former. Since $c' < b$, we know $b$ is not a lower bound of $\inf\{a \in [0, c] : (a, d) \cap P = \emptyset\}$. Thus, there exists some $a \in [0, c]$ with $a < b$ such that $(a, d) \cap P = \emptyset$. But then $a < b \le c < d$, so $b \in (a, d) \cap P = \emptyset$. This is a contradiction. the $d \le b < d'$ case can be proved analogously. Either way, $(c', d') \cap P = \emptyset$.
(Note: the above two paragraphs could be replaced by a nice Zorn's lemma argument to show a maximal open interval in $[0, 1] \setminus P$ exists.)
Now, choose $c'' \in [0, c') \cap P$ and $d'' \in (d', 1] \cap P$ such that \begin{align*} 0 &\le c' - c'' < \min\{1 - \beta, \beta\}(d' - c') \\ 0 &\le d'' - d' < \min\{1 - \beta, \beta\}(d' - c') \end{align*} This should be possible to do by construction of $c', d'$. Since $c'', d'' \in P$, we also know that $\beta c'' + (1 - \beta)d'' \in P$. I claim that this number lies in $(c', d')$, which is a contradiction! We have, \begin{align*} \beta c'' + (1 - \beta)d'' - c' &= (1 - \beta)(d'' - c'') - (c' - c'') \\ &> (1 - \beta)(d' - c') - \min\{1 - \beta, \beta\}(d' - c') \ge 0. \end{align*} We can similarly show that $d' - (\beta c'' + (1 - \beta)d'') > 0$. So, $$\beta c'' + (1 - \beta)d'' \in P \cap (c', d') = \emptyset,$$ a contradiction. Thus, $P$ must be dense.
(This came out longer than I expected. It may be easier just to find a closed form for $P$, as is done in the $\beta = 1/2$ case.)
