# Show that any convex combination of $x_1,...,x_k\in K$ is contained in $K$.

Assume that $$K\subseteq \mathbb{R}^n$$ is convex and $$x_1,...,x_k\in K$$. Show that any convex combination of $$x_1,...,x_k$$ is contained in $$K$$.

Relevant definitions:

Definition of Convexity: A set $$k\subseteq\mathbb{R}$$ is convex if for all $$x,y\in k$$, the line segment between $$x$$ and $$y$$ is entirely contained in $$k$$.

Definition of convex combination: A convex combination of $$x_1,...,x_k\in\mathbb{R}^n$$ is $$x=\lambda_1x_1+\lambda_2x_2+...+\lambda_kx_k$$ such that $$0\leq\lambda_i\leq1$$ and $$\lambda_1+\lambda_2+...+\lambda_k=1$$

My Attempt:

We can solve this by induction on $$k$$.

Base Case: By the definition of convexity, we know that the statement is true for $$k=2$$.

Induction Hypothesis: Assume that the statement holds for some value $$l$$. i.e. For $$x_1,...,x_l\in K$$, and $$\lambda_1,...,\lambda_l\geq0$$ with $$\lambda_1+...+\lambda_l=1$$, we have $$\lambda_1x_1+...+\lambda_lx_l\in K$$.

Induction Step: Let $$x_1,...x_l,x_{l+1}\in K$$ and $$\lambda_1,...,\lambda_l,\lambda_{l+1}\geq0$$ with $$\lambda_1+...+\lambda_l+\lambda_{l+1}=1$$. Now, assume that $$\lambda_1\neq 1$$. Then $$y=\lambda_1x_1+(1-\lambda_1)(\mu_2x_2+...+\mu_lx_l+\mu_{l+1}x_{l+1})$$ where $$\mu_i=\frac{\lambda_i}{1-\lambda_1}\geq0$$. Then $$\mu_2+...+\mu_{l+1}=\frac{\lambda_2+...+\lambda_{l+1}}{1-\lambda_1}=1$$ Since $$K$$ is convex and $$x_2,...x_{l+1}\in K$$, we have that $$\mu_2+,...\mu_{l+1}x_{l+1}\in K$$. Then since $$x_1\in K$$, we have that $$y\in K$$

I don't think I did this correctly because I couldn't find where I needed to apply the induction hypothesis. Any pointers would be very helpful. Thanks.

You did use the induction hypothesis to conclude that $$\sum_{i=2}^{l+1} \mu_i x_i \in K$$.
There are $$l$$ elements and you have just shown that $$\mu_2 + \ldots + \mu_{l+1}=1$$ and they are nonnegative and hence you can use the induction hypothesis.