## real analysis – A non-differentiable function \$f(x,y)\$ with bounded \$f_x\$, \$f_y\$, \$f_{xx}\$ and \$f_{yy}\$

Recently I was trying to construct a counterexample to the statement “If there exist $$f_{xy}(0,0)$$, $$f_{yx}(0,0)$$ and the functions $$f_{xx}$$, $$f_{yy}$$ exist in some neighborhood and are continuous at $$(0,0)$$, then $$f$$ is twice differentiable at $$(0,0)$$“. In order to do that the following question arose:

Is there a function $$fcolon mathbb{R}^2tomathbb{R}$$ with bounded $$f_x$$, $$f_y$$, $$f_{xx}$$, $$f_{yy}$$, which is non-differentiable at $$(0,0)$$?

If there exists such a function, then we’re done. The functions which came to mind have unbounded second partials $$f_{xx}$$ and $$f_{yy}$$.

## Let \$F\$ be a field. Show that \$F[x]/(x)\$ is isomorphic to \$F\$.

Can only help how to prove that. Let $$F$$ be a field Show that $$F(x)/(x)$$ is isomorphic to $$F$$.

## urr = fxx cos ^ 2x + 2fxy cosx sinx + fyy sin ^ 2 x

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For this question applies, if Ur should take place after the first partial derivation. It will result in a function Ur (x, y, θ). Why do I have to consider the following Urr values ​​the second time I make a partial derivative: Urr = ∂Ur / ∂x * ∂x / r + ∂ Ur / ∂y * ∂y / ∂r + ∂Ur / ∂θ * ∂θ / ∂r? Instead, the answer was Urr = ∂Ur / ∂x * ∂x / ∂r + ∂Ur / ∂y * ∂y / ∂r. Do not we have to take the last term into account? and if so, how would I set up a function to perform the partial derivative of ∂θ / ∂r?
Many Thanks