Let's carefully analyze the function and find its gradient, then verify its Fréchet differentiability. --- ### 1. **Given function:** \[ f : \mathbb{R}^n \to \mathbb{R}, \quad f(x) = \|x\|^2 \cdot 2e^{-\|x\|} \] where \(\|x\|\) denotes the Euclidean norm of \(x \in \mathbb{R}^n\). --- ### 2. **Rewrite the function:** Define \(r = \|x\|\). Then, \[ f(x) = 2 r^2 e^{-r} \] Note that \(f\) depends only on \(r\), which is a scalar function of \(x\). --- ### 3. **Find the gradient \(\nabla f(x)\):** Since \(f\) depends on \(x\) only through \(r = \|x\|\), we can use the chain rule: \[ \nabla f(x) = \frac{d f}{d r} \cdot \nabla r \] - First, compute \(\frac{d f}{d r}\): \[ f(r) = 2 r^2 e^{-r} \] \[ \frac{d f}{d r} = 2 \left( 2 r e^{-r} - r^2 e^{-r} \right) = 2 e^{-r} (2 r - r^2) \] - Next, compute \(\nabla r\): \[ r = \|x\| = \sqrt{\sum_{i=1}^n x_i^2} \] \[ \nabla r = \frac{x}{\|x\|} \quad \text{for } x \neq 0 \] At \(x=0\), the gradient of \(r\) is undefined, but in the context of differentiability, the function is smooth away from zero and can be extended at zero. --- ### 4. **Express the gradient:** Putting it all together, \[ \boxed{ \nabla f(x) = \frac{d f}{d r} \cdot \frac{x}{r} = 2 e^{-r} (2 r - r^2) \frac{x}{r} } \] for \(x \neq 0\). Simplify: \[ \nabla f(x) = 2 e^{-r} (2 r - r^2) \frac{x}{r} \] or equivalently, \[ \boxed{ \nabla f(x) = 2 e^{-\|x\|} \left( 2 \|x\| - \|x\|^2 \right) \frac{x}{\|x\|} \quad \text{for } x \neq 0 } \] --- ### 5. **Behavior at \(x=0\):** At \(x=0\), \(r=0\), \[ f(0) = 0 \] and the gradient at 0 can be computed as the limit: \[ \lim_{x \to 0} \nabla f(x) = \lim_{r \to 0} 2 e^{-r} (2 r - r^2) \frac{x}{r} \] Since \(x / r\) approaches any unit vector as \(x \to 0\), and \(2 r - r^2 \to 0\), the entire expression tends to zero. Therefore, \[ \nabla f(0) = 0 \] --- ### 6. **Summary of the gradient:** \[ \boxed{ \nabla f(x) = \begin{cases} 2 e^{-\|x\|} \left( 2 \|x\| - \|x\|^2 \right) \frac{x}{\|x\|}, & x \neq 0 \\ 0, & x=0 \end{cases} } \] --- ### 7. **Show that \(f\) is Fréchet differentiable on \(\mathbb{R}^n\):** - The function \(f\) is composed of smooth functions: - The norm \(\|x\|\) is smooth everywhere except at \(x=0\). But the composition \(f(r) = 2 r^2 e^{-r}\) is smooth in \(r \ge 0\). - \(f\) is radially symmetric and smooth for \(x \neq 0\). - At \(x=0\), the function is well-defined and continuous, and the gradient extends continuously to zero. - Since the gradient \(\nabla f(x)\) exists everywhere (including at \(x=0\)) and is continuous, \(f\) is **Fréchet differentiable** on all of \(\mathbb{R}^n\). - **Explicitly**, the Fréchet derivative at \(x\) is the linear map: \[ L_x(h) = \nabla f(x) \cdot h \] which depends continuously on \(x\). Hence, \(f\) is Fréchet differentiable everywhere. --- ### **### Final conclusion:** - The **gradient** of \(f\) at any \(x \in \mathbb{R}^n\) is: \[ \boxed{ \nabla f(x) = 2 e^{-\|x\|} (2 \|x\| - \|x\|^2) \frac{x}{\|x\|} \quad \text{for } x \neq 0, \quad \text{and } \nabla f(0) = 0 } \] - The function \(f\) is **Fréchet differentiable** on \(\mathbb{R}^n\), with the derivative given by the above gradient, which is continuous everywhere. --- **This completes the step-by-step derivation and proof.**

Let's solve the problem step by step. --- ## **1. The function** Given: \[ f:\mathbb{R}^n \to \mathbb{R}, \qquad f(x) = \|x\|^2 e^{-\|x\|} \] where \(\|x\|\) is the Euclidean norm. --- ## **2. Compute the gradient \(\nabla f(x)\)** ### **Step 1: Write \(f(x)\) in terms of \(r = \|x\|\)** \[ f(x) = r^2 e^{-r} \] where \(r = \|x\|\). ### **Step 2: Use the chain rule** The gradient is: \[ \nabla f(x) = \frac{df}{dr} \cdot \nabla r \] #### **a. Compute \(\frac{df}{dr}\):** \[ \frac{d}{dr}(r^2 e^{-r}) = 2r e^{-r} + r^2(-e^{-r}) = e^{-r}(2r - r^2) \] #### **b. Compute \(\nabla r\):** \[ r = \|x\| = \sqrt{x_1^2+\cdots+x_n^2} \] \[ \frac{\partial r}{\partial x_k} = \frac{x_k}{\|x\|} \] So, \[ \nabla r = \frac{x}{\|x\|} \quad (x \neq 0) \] #### **c. Multiply:** \[ \nabla f(x) = e^{-r}(2r - r^2)\cdot \frac{x}{r} \] \[ = e^{-\|x\|} (2\|x\| - \|x\|^2) \frac{x}{\|x\|}, \quad x \neq 0 \] --- ### **Step 3: Simplify and check at \(x=0\)** When \(x = 0\): - \(f(0) = 0\) - For the gradient, consider the limit as \(x\to 0\): \[ e^{-\|x\|}(2\|x\| - \|x\|^2)\frac{x}{\|x\|} \to 0 \] So, define \(\nabla f(0) = 0\). --- ### **Step 4: Final formula** \[ \boxed{ \nabla f(x) = \begin{cases} e^{-\|x\|} (2\|x\| - \|x\|^2) \dfrac{x}{\|x\|}, & x \neq 0 \\ 0, & x = 0 \end{cases} } \] --- ## **3. Fréchet differentiability** ### **Step 1: Is \(f\) differentiable everywhere?** - For \(x \neq 0\): All terms are smooth compositions of smooth functions, so \(\nabla f\) exists and is continuous. - At \(x = 0\): We've shown \(\nabla f(0) = 0\). **To check Fréchet differentiability at \(x=0\):** Recall: \(f\) is Fréchet differentiable at \(x\) if \[ \lim_{h \to 0} \frac{f(x+h)-f(x) - \nabla f(x)\cdot h}{\|h\|} = 0 \] Take \(x = 0\): \[ f(h) = \|h\|^2 e^{-\|h\|} \] \[ \nabla f(0) = 0 \] So, \[ \frac{f(h) - 0 - 0}{\|h\|} = \|h\| e^{-\|h\|} \] As \(h \to 0\), \(\|h\| \to 0\), \(e^{-\|h\|} \to 1\), so the expression \(\|h\| e^{-\|h\|} \to 0\). **So the limit is zero, and \(f\) is Fréchet differentiable at \(0\).** --- ### **Step 2: Is the gradient continuous?** - \(\nabla f(x)\) is continuous everywhere (including at \(x=0\)), since as \(x \to 0\), \[ e^{-\|x\|} (2\|x\| - \|x\|^2) \frac{x}{\|x\|} \to 0 \] So the derivative is continuous. --- ## **4. Conclusion** - The **gradient** for all \(x\) is \[ \boxed{ \nabla f(x) = \begin{cases} e^{-\|x\|} (2\|x\| - \|x\|^2) \dfrac{x}{\|x\|}, & x \neq 0 \\ 0, & x = 0 \end{cases} } \] - The function \(f\) is **Fréchet differentiable everywhere** on \(\mathbb{R}^n\), with the derivative given above. --- **Summary:** We computed the gradient step by step using the chain rule and showed that \(f\) is Fréchet differentiable everywhere on \(\mathbb{R}^n\), with continuous gradient.