Let $x\in\mathbb{R}^n$ be a centered random vector (i.e, $\mathbf{E}[x]=0$), and $u\in \mathbb{S}^{n-1}$ be an arbitrary unit vector. Use "$\cdot$" to represent inner product.
If the variance $\mathbf{E}[x\cdot x]\leq \sigma^2$, then we have $\mathbf{E}[(x\cdot u)^2]\leq \sigma^2$ for any $u$. This is simply due to the Cauchy-Schwarz inequality, as $\mathbf{E}[(x\cdot u)^2]\leq \mathbf{E}[(x\cdot x)(u\cdot u)]=\mathbf{E}[x\cdot x]\leq \sigma^2$.
I am wondering whether the converse is true. That is, if we assume that $\mathbf{E}[(x\cdot u)^2]\leq \sigma^2$ holds for any unit vector $u$, can we conclude that $\mathbf{E}[x\cdot x]\leq \sigma^2$? If not, I will appreciate it if someone can help me find a counter-example.
(Of course, we can find $\mathbf{E}[x\cdot x]\leq n\sigma^2$, but I am wondering whether there is better result.)
Well, this question seems a little stupid now.
Let $n=2$ and consider the random vector $x=(X,Y)$, where $X$ and $Y$ are independent standard normal random variables.
Use $u=(\cos\theta,\sin\theta)$ to represent an arbitrary unit vector. Then, $x\cdot u=X\cos\theta+Y\sin\theta$ is also a standard normal random variable, and $\mathbf{E}[(x\cdot u)^2]=1$ for any unit vector $u$.
However, $\mathbf{E}[x\cdot x]=\mathbf{E}[X^2+Y^2]=2$. Hence, the converse does not hold in general. This kind of counterexample can also be constructed similarly for higher dimensions.
It turns out that the implication in both directions does not really match. If the variance $\mathbf{E}[x\cdot x]\leq\sigma^2$, we can actually get better conclusion than $\mathbf{E}[(x\cdot u)^2]\leq \sigma^2$. Let $\{u_1, u_2, \dots, u_n\}$ be an orthogonal basis, then we have $\mathbf{E}[x\cdot x]= \mathbf{E}[ \{\sum_{i=1}^n (u_i\cdot x)u_i\}\cdot \{\sum_{j=1}^n (u_j\cdot x)u_j\}] = \sum_{i=1}^n \mathbf{E}[ (u_i\cdot x)^2 ]\leq \sigma^2. $
That is, if we define $\sigma_i^2 = \mathbf{E}[ (u_i\cdot x)^2 ]$, then $\sum_{i=1} \sigma_i^2\leq \sigma^2$.