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 $\lambda=\frac{1}{2}$. 



Is duality of convex bodies in itself a convex function?
Does this work? 


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.


