Let $Sec\_r(V)$ be the $r$-secant variety of a Veronse variety $V\subset\mathbb{P}^N$, that is 
$$Sec\_r(V) = \bigcup\_{p\_1,...,p\_r\in V}\left\langle p\_1,...,p\_r\right\rangle\subset\mathbb{P}^N$$
where $V$ is the image of $\mathbb{P}^n$ via the embedding induced by $\mathcal{O}\_{\mathbb{P}^n}(d)$.


Is it true that if $Sec\_r(V)\neq\mathbb{P}^N$ then any polynomial on $\mathbb{P}^N$ vanishing on $Sec\_r(V)$ has degree greater or equal than $r+1$ ? 



Degree of equations of secant varieties of Veronese varieties
The answer by JM Landsberg in the link from aginensky's comment precisely says that the $r+1$ lower bound on the degree of generators is true.
See top of page 2 in the article ["Prolongations and computational algebra"](https://arxiv.org/abs/math/0611696) by Sidman and Sullivant where they point to the article ["On the ideal of an embedded join"](http://www.sciencedirect.com/science/article/pii/S0021869399980912) by Ulrich and Simis. 




---


Edit: Having thought about the question a bit more, I realized this is trivial (for the Veronese, but not for secants of general varieties as in the above articles) if one knows the symbolic method from 19th century invariant theory.


Consider the secant $\sigma\_r(v\_d(\mathbb{P}^n))$ for the degree $d$ Veronese embedding. Let $C(F)$ be a homogeneous polynomial of degree $r$ in the coefficients of a generic form $F$ of degree $d$ in $n+1$ variables.
If $C(F)=0$ for all $F$'s which can be written $F=L\_1^d+\cdots +L\_r^d$ for some linear forms $L\_i$, then
the polynomial $C$ vanishes identically.


Indeed,
$$
C(F)=\left.M(F\_1,\ldots,F\_r)\right|\_{F\_1=\cdots=F\_r=F}
$$
where $M$ is the associated symmetric multilinear form given by
$$
M(F\_1,\ldots,F\_r)=\frac{1}{r!}\frac{\partial^r}{\partial t\_1\cdots \partial t\_r}
\ C(t\_1 F\_1+\cdots+t\_r F\_r)\ .
$$
From the hypothesis (and working over a field like $\mathbb{C}$ which contains $d$-th roots of unity) one immediately gets
$$
M(L\_1^d,\ldots,L\_r^d)=0
$$
for linear all forms $L\_1,\ldots,L\_r$.


Finally, one has the identity 
$$
C(F)=\frac{1}{d!^r}\ F(\frac{\partial}{\partial L\_1})\cdots
F(\frac{\partial}{\partial L\_r})\ 
M(L\_1^d,\ldots,L\_r^d)
$$
where the point coordinates $x\_1,\ldots,x\_{n+1}$ are replaced by differential operators in the coefficients of the linear forms $L\_1,\ldots,L\_r$. As a result, $C(F)$ is identically zero.
The last identity is what makes the classical symbolic method work. For more on this, please see [this article](http://www.sciencedirect.com/science/article/pii/S0021869315005657) (or [here](https://arxiv.org/abs/1402.2669) for the preprint version).


