Proving |f(x)-f(y)| ≤ K ||x-y|| for f:R^2→R

[onie mti]

Suppose that f:R^2 to R is differentiable on R^{2}. Also assume that there exists a real number K(greater that or equal to) 0, so that 2-norm of the (gradient of (f(x)) )is less than or equal to K for all x,y in R^{2}. Prove that |f(x)-f(y)| is less than or equal to K(multiply by the 2-norm of x-y) for all x,y in R^2.

i tried applying the mean value theorem to the function g(t)= f((1-t)x+ty) t is in [0,1] but I cannot move forward

it is no 2 on the uploaded files

 

[Opalg]

Re: converging maps

Suppose that f:R^2 to R is differentiable on R^{2}. Also assume that there exists a real number K(greater that or equal to) 0, so that 2-norm of the (gradient of (f(x)) )is less than or equal to K for all x,y in R^{2}. Prove that |f(x)-f(y)| is less than or equal to K(multiply by the 2-norm of x-y) for all x,y in R^2.

i tried applying the mean value theorem to the function g(t)= f((1-t)x+ty) t is in [0,1] but I cannot move forward

it is no 2 on the uploaded files

Hi onie mti, and welcome to MHB! Suppose you define a function $k:\mathbb{R}\to \mathbb{R}^2$ by $k(t) = (1-t)\mathbf{x} + t\mathbf{y}$, with derivative $k'(t) = \mathbf{y} - \mathbf{x}$. Then $g$ is the composition $f\circ k$, and the chain rule says that $g'(t) = \nabla f\bigl(k(t)\bigr)\cdot k'(t)$. Now apply the Cauchy–Schwarz inequality to see that $|g'(t)| \leqslant \|\nabla f\bigl(k(t)\bigr)\|_2\| k'(t)\|_2$.

 

[onie mti]

Re: converging maps

Hi onie mti, and welcome to MHB! Suppose you define a function $k:\mathbb{R}\to \mathbb{R}^2$ by $k(t) = (1-t)\mathbf{x} + t\mathbf{y}$, with derivative $k'(t) = \mathbf{y} - \mathbf{x}$. Then $g$ is the composition $f\circ k$, and the chain rule says that $g'(t) = \nabla f\bigl(k(t)\bigr)\cdot k'(t)$. Now apply the Cauchy–Schwarz inequality to see that $|g'(t)| \leqslant \|\nabla f\bigl(k(t)\bigr)\|_2\| k'(t)\|_2$.

thank you :) let me get on it (Bow)

 

[onie mti]

Re: converging maps

Hi onie mti, and welcome to MHB! Suppose you define a function $k:\mathbb{R}\to \mathbb{R}^2$ by $k(t) = (1-t)\mathbf{x} + t\mathbf{y}$, with derivative $k'(t) = \mathbf{y} - \mathbf{x}$. Then $g$ is the composition $f\circ k$, and the chain rule says that $g'(t) = \nabla f\bigl(k(t)\bigr)\cdot k'(t)$. Now apply the Cauchy–Schwarz inequality to see that $|g'(t)| \leqslant \|\nabla f\bigl(k(t)\bigr)\|_2\| k'(t)\|_2$.

i managed to use the schwarz inequality inequality but i can not show that |f(x)-f(y)| is less than or equal to K(multiply by the 2-norm of x-y)

 

[Opalg]

Re: converging maps

Hi onie mti, and welcome to MHB! Suppose you define a function $k:\mathbb{R}\to \mathbb{R}^2$ by $k(t) = (1-t)\mathbf{x} + t\mathbf{y}$, with derivative $k'(t) = \mathbf{y} - \mathbf{x}$. Then $g$ is the composition $f\circ k$, and the chain rule says that $g'(t) = \nabla f\bigl(k(t)\bigr)\cdot k'(t)$. Now apply the Cauchy–Schwarz inequality to see that $|g'(t)| \leqslant \|\nabla f\bigl(k(t)\bigr)\|_2\| k'(t)\|_2$.

i managed to use the schwarz inequality inequality but i can not show that |f(x)-f(y)| is less than or equal to K(multiply by the 2-norm of x-y)

Have you used the hint about the mean value theorem? It says that $g(1) - g(0) = g'(t)$ for some $t$. Then $$|f(\mathbf{y}) - f(\mathbf{x})| = |g(1) - g(0)| = |g'(t) \leqslant \|\nabla f\bigl(k(t)\bigr)\|_2\| k'(t)\|_2 \leqslant K\| \mathbf{y} - \mathbf{x}\|_2.$$