Assume $\langle v,s \rangle + \langle s,v \rangle \leq \langle s,s \rangle$

The question is

Let $$V$$ be a complex inner product space, and let $$S$$ be a subspace of $$V$$. Suppose that $$v\in V$$ is a vector for which $$\langle s,v\rangle + \langle v,s\rangle \leq \langle s,s\rangle$$ for all $$s\in S$$. Prove that $$v\in S^{\perp}$$.

I am thinking about proving it by contradiction, but I am not sure what $$\langle s,v\rangle + \langle v,s\rangle \leq \langle s,s\rangle$$ can tell me. What I am sure about right now is that $$v$$ must not be in $$S$$ since if $$v$$ is in $$S$$, then $$v$$ will be equal to some $$s$$ in $$S$$, then there exists such $$s$$ that $$\langle s,v\rangle + \langle v,s\rangle = \langle s,s\rangle + \langle s,s\rangle = 2\langle s,s\rangle \gt \langle s,s\rangle$$. So $$v$$ must be in somewhere else. What else can I know, I am so confused right now, can somebody give me some hints?

• As a side comment, you need to assume $s\neq \mathbf{0}$ in your bit of argument. Otherwise, $2\langle s,s\rangle$ could equal $\langle s,s\rangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^{“\perp}$. – Arturo Magidin Mar 14 at 4:53
• Is $S$ finite dimensional? If so, write $v=s+p$, where $s\in S$ and $P\in S^{\perp}$. – Arturo Magidin Mar 14 at 4:55
• @ArturoMagidin the question does not provide it is finite dimensional or not. – PixieBlade Mar 14 at 4:58
• @ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S. – miracle173 Mar 14 at 5:08
• Fix nonzero $s\in S$ (the case $S = 0$ is trivial), and consider $\lambda s\in S$ for small $\lambda$. – anomaly Mar 14 at 5:19

Fix $$s \in S$$. Replace $$s$$ by $$\epsilon s$$ and divide by $$\epsilon$$ to get $$\langle v, s \rangle +\langle s, v \rangle \leq \epsilon \|s\|^{2}$$. Letting $$\epsilon \to 0$$ we see that Real part of $$\langle v, s \rangle$$ is $$\leq 0$$. Replace $$s$$ by $$-s$$ to see that the real part is $$0$$. Replace $$s$$ by $$is$$ to see that the imaginary part is also $$0$$.

Hint: Think about what happens as we make $$\langle s,s\rangle$$ smaller and smaller.

• Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking... – PixieBlade Mar 14 at 5:02
• Yep, you're on the right track! You can rigorize this by taking a vector $\vec{v}_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector. – Isaac Browne Mar 14 at 5:05

If $$S\ne\{0\}$$ we can choose an $$s_1\ne 0$$ and that means $$\langle s_1,s_1\rangle\ne 0$$ We set $$s_2=\frac{1}{\sqrt{\langle s_1, s_1\rangle}}s_1$$ then we have $$\langle s_2, s_2 \rangle=1$$

For an arbitrary $$v \in V$$ we set $$s=\langle v,s_2\rangle s_2 \in S$$

and for the LHS of

$$\langle s,v\rangle + \langle v,s\rangle \leq \langle s,s\rangle$$ we get

$$\langle s,v\rangle + \langle v,s\rangle \\=\langle \langle v,s_2\rangle s_2,v\rangle + \langle v,\langle v,s_2\rangle s_2\rangle\\=\langle v,s_2\rangle \langle s_2,v\rangle+\overline{\langle v,s_2\rangle }\langle v,s_2\rangle\\=2\overline{\langle v,s_2\rangle }\langle v,s_2\rangle$$

For the RHS we get $$\langle s ,s \rangle\\=\langle \langle v,s_2\rangle s_2,\langle v,s_2\rangle s_2 \rangle\\=\langle v,s_2\rangle \overline{\langle v,s_2\rangle}\langle s_2,s_2 \rangle=\\\langle v,s_2\rangle \overline{\langle v,s_2\rangle}$$

It follows that $$2\overline{\langle v,s_2\rangle }\langle v,s_2\rangle \le \overline{\langle v,s_2\rangle }\langle v,s_2\rangle$$ which is a contradiction for $$\overline{\langle v,s_2\rangle }\langle v,s_2\rangle\ne 0$$