# Is duality of convex bodies in itself a convex function?

Let $K,T\subset\mathbb{R}^{n}$ be convex bodies (i.e. nonempty, compact and convex) containing the origin, and let $\lambda\in(0,1)$. Fix $\langle \cdot ,\cdot \rangle$ to be the standard dot product in $\mathbb{R}^n$ (though probably here any dot product would do). Let us denote by $K^\circ$ the dual to $K$ with respect to the fixed dot product, and recall that $K^\circ=\{y\in\mathbb{R}^n : \langle x, y\rangle \leq 1, \forall x\in K\}$.

Is it true that $$(\lambda K + (1-\lambda) T)^\circ\subset \lambda K^\circ + (1-\lambda)T^\circ?$$

So far I managed to show the equivalence of this question to the following: Denote by $h_K(x)=\sup_{y\in K} \langle x, y \rangle$ the support function of $K$, and by $g_K(x)=\inf \{ r>0: x\in rK\}$ the gauge function of $K$. By some very basic manipulations one can see that the original question is equivalent to asking whether or not it's true that $$h_{(\lambda K + (1-\lambda)T)^\circ}\leq \lambda h_{K^\circ} + (1-\lambda)h_{T^\circ}.$$ This is already quite believable. Using the fact that for convex bodies (as above) it holds that $h_{K^\circ}=g_K$, and that for such bodies $g_K$ is nearly a norm (i.e. it's sublinear, positively homogeneous and non-negative), then one obtains $$\Vert\cdot \Vert_{\lambda K +(1-\lambda)T} \leq \lambda\Vert\cdot\Vert_K + (1-\lambda)\Vert\cdot\Vert_T.$$ I don't know if the last inequality is true, but as far as I understand convexity, it has every right to be true. Is there a functional analysis approach I might be missing?

Lastly, if the above is not true in general I would be satisfied in answering this for $\lambda=\frac{1}{2}$.

First prove it when $n=1$. Now reduce the general case to this one as follows:
If $L$ is a line through the origin in $\mathbb R^n$ and $K$ is convex in $\mathbb R^n$ then the intersection of the dual of $K$ with $L$ is the dual (with respect to $L$) of the image of $K$ in $L$ under orthogonal projection.
\begin{align*} h_{((1-\lambda)K + \lambda T)^\circ}(x) &= \sup_\xi \frac{\langle\xi,x\rangle}{h_{(1-\lambda)K + \lambda T}(\xi)}\\ &= \sup_\xi \frac{\langle\xi,x\rangle}{(1-\lambda)h_K(\xi) + \lambda h_T(\xi)}\\ &\le \sup_\xi \left[(1-\lambda)\frac{\langle\xi,x\rangle}{h_K(\xi)} + \lambda\frac{\xi,x\rangle}{h_T(\xi)}\right]\\ &\le (1-\lambda)h_{K^\circ}(x) + \lambda h_{T^\circ}(x)\\ &= h_{(1-\lambda)K^\circ+\lambda T^\circ}(x) \end{align*} Therefore, $$((1-\lambda)K + \lambda T)^{\circ} \subset (1-\lambda)K^\circ + \lambda T^\circ.$$ Equality holds if and only if $T = tK$ for some $t > 0$.