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?

  • $\begingroup$ 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 $S^{“\perp}$. $\endgroup$ – Arturo Magidin Mar 14 at 4:53
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    $\begingroup$ Is $S$ finite dimensional? If so, write $v=s+p$, where $s\in S$ and $P\in S^{\perp}$. $\endgroup$ – Arturo Magidin Mar 14 at 4:55
  • $\begingroup$ @ArturoMagidin the question does not provide it is finite dimensional or not. $\endgroup$ – PixieBlade Mar 14 at 4:58
  • $\begingroup$ @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. $\endgroup$ – miracle173 Mar 14 at 5:08
  • $\begingroup$ Fix nonzero $s\in S$ (the case $S = 0$ is trivial), and consider $\lambda s\in S$ for small $\lambda$. $\endgroup$ – 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.

  • $\begingroup$ 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... $\endgroup$ – PixieBlade Mar 14 at 5:02
  • $\begingroup$ 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. $\endgroup$ – 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 $$\overline{\langle v,s_2\rangle }\langle v,s_2\rangle\ne 0$$


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