# Confusion about the convexity of the best response correspondence

I am recently reading the proof of the existence of the Nash Equilibrium. As a math student, I do understand the use of Berge's maximum theorem and Kakutani's fixed point theorem, but I am not sure why the mixed-strategy space and the image of the best response correspondence are convex.

For example, if I have two mixed-strategies $$\sigma_i,\sigma_i'\in BR_i(\sigma_{-i})$$, why is the mix of these two strategies, $$\lambda\sigma_i+(1-\lambda)\sigma_i'$$ ($$0\leqslant\lambda\leqslant1$$), also a best response? Since $$\sigma_i,\sigma_i'\in BR_i(\sigma_{-i})$$ suggests $$u_i(\sigma_i,\sigma_{-i})=u_i(\sigma’_i,\sigma_{-i})=\max\{u_i(x_i,\sigma_{-i})\}$$, how to prove that $$u_i(\lambda\sigma_i+(1-\lambda)\sigma_i',\sigma_{-i})$$ also equals to maximum?

I would have thought that the if the opponent strategy $$\sigma_{-i}$$ is given, even if it is a mixed strategy, then $$u_i(\lambda\sigma_i+(1-\lambda)\sigma_i', \sigma_{-i})= \lambda u_i(\sigma_i, \sigma_{-i})+(1-\lambda)u_i(\sigma_i', \sigma_{-i})$$ and so if $$u_i(\sigma_i, \sigma_{-i})=u_i(\sigma_i', \sigma_{-i})$$ then $$u_i(\lambda\sigma_i+(1-\lambda)\sigma_i', \sigma_{-i})$$ is equal to both of them. If they are best responses then so too is the linear combination strategy