If $f:\{\lambda a+(1-\lambda)b:0\leq\lambda \leq 1\}\to \mathbb{R}$ is a linear function, then is it true that $f$ is continuous?

Question

Question: Let $V$ be a real vector space and $a,b\in V$ such that $a\neq b.$ If $f:\{\lambda a+(1-\lambda)b:0\leq\lambda \leq 1\}\to \mathbb{R}$ is a linear function, then is it true that $f$ is continuous?

I think it is true. Since $f$ is linear, it suffices to show that $f$ is bounded. Let $M = \max\{|f(a)|,|f(b)|\}.$ Note that $$|f(\lambda a +(1-\lambda)b)| = |\lambda f(a) + (1-\lambda) f(b)| \leq \lambda M + (1-\lambda)M = M.$$ So $f$ is bounded.

Is my proof above correct?

Answer 1

Let $x=\lambda_{x}a+(1-\lambda_{x})b$ and $y=\lambda_{y}a+(1-\lambda_{y})b$ be points in $\operatorname{dom}f$. Note that $$ x-y=\left(\lambda_{x}-\lambda_{y}\right)\left(a-b\right). $$ Suppose $V$ is a normed vector space and that $a\neq b$ (the case of $a=b$ is uninteresting since $\operatorname{dom}f$ becomes a singleton). By the above, $$ \left|\lambda_{x}-\lambda_{y}\right|=\frac{\left\Vert x-y\right\Vert }{\left\Vert a-b\right\Vert }. $$ Moreover, $$ \left|f(x)-f(y)\right|=\left|f(x-y)\right|=\left|\lambda_{x}-\lambda_{y}\right|\left|f(a-b)\right|, $$ and hence $f$ is continuous since $|\lambda_{x}-\lambda_{y}|\rightarrow0$ as $\left\Vert x-y\right\Vert \rightarrow0$.