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Let $\textbf a \in \mathbb R^n$ be a nonzero vector.Show that maximum of $\textbf a^{ T} \textbf x$ over $ B[0,1]=\{\textbf x \in \mathbb R^n: \lVert \textbf x \rVert \le 1 \}$ is attained at $\textbf x^*=\frac{\textbf a}{\lVert \textbf a \rVert}$ and that the maximal value is $\lVert \textbf a \rVert$.

The function $f:\mathbb R^n \to \mathbb R$ with $\textbf x \mapsto \textbf a^{ T} \textbf x$ is continuous on $\mathbb R^n$, Moreover the set $ B[0,1]$ is closed and bounded. So on $B[0,1]$ the function $f$ does have both a global maximum and a global minimum.If $\textbf x \in B(0,1)$ is a global extremum of function $f$ over $ B[0,1]$ then $\nabla f(\textbf x )=\textbf a=\textbf 0 $ which is not possible since $\textbf a$ is a nonzero vector, Hence $\textbf x$ should be on the boundary of $ B[0,1]$.

This implies that $\lVert \textbf x \rVert=1$, I don't know how to proceed, It should be mentioned that I don't know about lagrange multiplier, so I won't use that.

user1040538
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1 Answers1

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For $x \in B[0,1]$ we have, by Cauchy - Schwarz:

$$|f(x)| \le ||a|| \cdot ||x|| \le ||a||.$$

Hence

$$ -||a|| \le f(x) \le ||a||.$$

Let $x_0:= \frac{a}{||a||}$ and look at $f(x_0)$ and $f(-x_0).$

Fred
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  • Yes, $f(x) \le ||a||$ and $f(x_0)=||a||.$ Furthermore we have $-||a|| \le f(x)$ and $f(x_1)=-||a||.$ – Fred May 04 '22 at 07:45