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Let $E$ and $F$ be (real) Hilbert spaces, where $\dim E = \infty$ and $1 \leq \dim F < \infty$. Let $T : E \to \operatorname{Sym}(F \times F,\mathbb{R})$ be a continuous linear map, where $\operatorname{Sym}(F \times F, \mathbb{R})$ denotes the space of symmetric bilinear maps $F \times F \to \mathbb{R}$. Suppose that for every $x \in E$ the quadratic form

$$F \ni v \mapsto T(x)(v,v) \in \mathbb{R}$$

takes both nonpositive and nonnegative values.

My question: is there a continuous/differentiable map $s : E \to F\setminus \{0\}$ such that $T(x)(s(x), s(x)) = 0$ for any $x \in E$? In other words, is it always possible to make a continuous choice of an isotropic direction for the bilinear forms $T(x)$ when $x$ varies on $E$?

I suspect it is possible, but I have no clue on how to show it.

Let $E$ and $F$ be (real) Hilbert spaces, where $\dim E = \infty$ and $1 \leq \dim F < \infty$. Let $T : E \to \operatorname{Sym}(F \times F,\mathbb{R})$ be a continuous linear map, where $\operatorname{Sym}(F \times F, \mathbb{R})$ denotes the space of symmetric bilinear maps $F \times F \to \mathbb{R}$. Suppose that for every $x \in E$ the quadratic form

$$F \ni v \mapsto T(x)(v,v) \in \mathbb{R}$$

takes both nonpositive and nonnegative values when calculated in nonzero vectors of $F$.

My question: is there a continuous/differentiable map $s : E \to F\setminus \{0\}$ such that $T(x)(s(x), s(x)) = 0$ for any $x \in E$? In other words, is it always possible to make a continuous choice of an isotropic direction for the bilinear forms $T(x)$ when $x$ varies on $E$?

I suspect it is possible, but I have no clue on how to show it.

Let $E$ and $F$ be (real) Hilbert spaces, where $\dim E = \infty$ and $2 \leq \dim F < \infty$. Let $T : E \to \operatorname{Sym}(F \times F,\mathbb{R})$ be a continuous linear map, where $\operatorname{Sym}(F \times F, \mathbb{R})$ denotes the space of symmetric bilinear maps $F \times F \to \mathbb{R}$. Suppose that for every $x \in E \setminus \{0\}$ the quadratic form

$$F \ni v \mapsto T(x)(v,v) \in \mathbb{R}$$

is indefinite, i.e., takes both positive and negative values.

My question: is there a continuous/differentiable map $s : E \to F\setminus \{0\}$ such that $T(x)(s(x), s(x)) = 0$ for any $x \in E$? In other words, is it always possible to make a continuous choice of an isotropic direction for the bilinear forms $T(x)$ when $x$ varies on $E$?

I suspect it is possible, but I have no clue on how to show it.

Let $E$ and $F$ be (real) Hilbert spaces, where $\dim E = \infty$ and $1 \leq \dim F < \infty$. Let $T : E \to \operatorname{Sym}(F \times F,\mathbb{R})$ be a continuous linear map, where $\operatorname{Sym}(F \times F, \mathbb{R})$ denotes the space of symmetric bilinear maps $F \times F \to \mathbb{R}$. Suppose that for every $x \in E$ the quadratic form

$$F \ni v \mapsto T(x)(v,v) \in \mathbb{R}$$

takes both nonpositive and nonnegative values.

My question: is there a continuous/differentiable map $s : E \to F\setminus \{0\}$ such that $T(x)(s(x), s(x)) = 0$ for any $x \in E$? In other words, is it always possible to make a continuous choice of an isotropic direction for the bilinear forms $T(x)$ when $x$ varies on $E$?

I suspect it is possible, but I have no clue on how to show it.

Let $E$ and $F$ be (real) Hilbert spaces, where $\dim E = \infty$ and $2 \leq \dim F < \infty$. Let $T : E \to \operatorname{Sym}(F \times F,\mathbb{R})$ be a continuous linear map, where $\operatorname{Sym}(F \times F, \mathbb{R})$ denotes the space of symmetric bilinear maps $F \times F \to \mathbb{R}$. Suppose that for every $x \in E$ the quadratic form

$$F \ni v \mapsto T(x)(v,v) \in \mathbb{R}$$

is indefinite, i.e., takes both positive and negative values.

My question: is there a continuous/differentiable map $s : E \to F\setminus \{0\}$ such that $T(x)(s(x), s(x)) = 0$ for any $x \in E$? In other words, is it always possible to make a continuous choice of an isotropic direction for the bilinear forms $T(x)$ when $x$ varies on $E$?

I suspect it is possible, but I have no clue on how to show it.

Let $E$ and $F$ be (real) Hilbert spaces, where $\dim E = \infty$ and $2 \leq \dim F < \infty$. Let $T : E \to \operatorname{Sym}(F \times F,\mathbb{R})$ be a continuous linear map, where $\operatorname{Sym}(F \times F, \mathbb{R})$ denotes the space of symmetric bilinear maps $F \times F \to \mathbb{R}$. Suppose that for every $x \in E \setminus \{0\}$ the quadratic form

$$F \ni v \mapsto T(x)(v,v) \in \mathbb{R}$$

is indefinite, i.e., takes both positive and negative values.

My question: is there a continuous/differentiable map $s : E \to F\setminus \{0\}$ such that $T(x)(s(x), s(x)) = 0$ for any $x \in E$? In other words, is it always possible to make a continuous choice of an isotropic direction for the bilinear forms $T(x)$ when $x$ varies on $E$?

I suspect it is possible, but I have no clue on how to show it.