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EDIT: To make sense, the function in the question must be some $f:[0,1]\times [0,1]\rightarrow \mathbb{R}$. Then, what I want to show is that $f$ is continuous with respect to its second argument. This is why I choose $\lambda$ equal to the sup (or max) of $|f(x)-f(x_0)|$, which does not depend on the first argument but it does depend on $x$ and $x_0$$

I am not sure about the use of $\lambda$ when negating continuity. Fix the domain of $f$. Suppose that $f$ is not continuous on $[0,1]$. Then, negating the def of continuity we get: $\exists \,x_0\in[0,1]$ $\exists \, \epsilon>0$ such that $\forall \lambda>0$ $\exists x\in [0,1]$ such that $|x-x_0|<\lambda \text{ and } |f(x)-f(x_0)|\geq \epsilon$. My understanding is that this negation should work for every $\lambda>0$ which allows me, for example, to choose $\lambda=\sup_{t}|f(x)-f(x_0)|$. My doubts is that this particular $\lambda$ depends on $x$ as well as on $x_0$ and I'm not sure this is correct.

Any help is greatly appreciated it!

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Why do both $t$ and $x$ occur in your formula? Anyway, the interesting part is that you can choose $\lambda$ as small as you want, so that using $\sup$ seems strange. Also, $\lambda$ is "on the $x$ axis" so that an expression involving $f(x_0)$ etc. seems even stranger. – Hagen von Eitzen Dec 30 '12 at 22:35
@HagenvonEitzen I edited the question to make sense. The thing is that I want $f$ to be continuous (pointwise) on its second argument, so my choice of $\lambda$ deals with the first argument. – Cristian Dec 30 '12 at 22:48
There is no second argument. You are dealing with a function of one variable, defined on $[0,1]$. – Robert Israel Dec 30 '12 at 22:53
@RobertIsrael if we fix $t\in[0,1]$ then we get a function of a single variable (check the edit of my question). However, my doubt about $\lambda$ and its role in the negation still remains valid... – Cristian Dec 30 '12 at 22:57
If you want to ask about a function of two variables, do so. Right now it's very hard to tell what exactly is the question. – Robert Israel Dec 30 '12 at 23:00
up vote 0 down vote accepted

If $f$ is not continuous on $[0,1]$, it is discontinuous at some $x_0 \in [0,1]$, and there is some $\epsilon > 0$ such that it is impossible to ensure that $|f(x) - f(x_0)| < \epsilon$ by making $x$ be close to $x_0$. That is, for every $\lambda > 0$ there is some $x$ with $|x - x_0| < \lambda$ and $|f(x) - f(x_0)| > \epsilon$. But that $x$ will certainly have to depend on $\lambda$: any particular $x \ne x_0$ won't work if $\lambda \le |x - x_0|$. It makes no sense to have $\lambda$ dependent on $x$.

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