Halting problem with extra input

Question

Can there be a function HALT(f, y) so that:

• There are some x such that f(x) halts iff there are some y such that HALT(f, y) return true;
• There are some x such that f(x) doesn't halt iff there are some y such that HALT(f, y) return false;
• HALT always halt and return a boolean

? It makes lots of old conflict don't work. E.g. If f(x) loops forever when HALT(f, x) is true, HALT may return true for half of x.

Answer 1

I assume you mean "Turing machine" by "function", otherwise it is meaningless to say a function halts on some input.

Suppose $\mathrm{HALT}$ exists, then we construct a Turing machine $H$ as follows:

On input $\langle M, w\rangle$ where $M$ is an (encoding of) Turing machine:

1. Construct a Turing machine $N_{\langle M, w\rangle}$ with input $x$ as follows: run $M$ on $x$ and return what $M$ returns.

2. Run $\mathrm{HALT}$ on $\langle N_{\langle M, w\rangle}, 0\rangle$ and return what $\mathrm{HALT}$ returns.

Note the result of running $N_{\langle M, w\rangle}$ has nothing to do with its input $x$. If $M$ halts on $w$, $N_{\langle M, w\rangle}$ halts on all inputs, thus $\mathrm{HALT}( N_{\langle M, w\rangle}, 0)$ returns true. otherwise $N_{\langle M, w\rangle}$ halts on no input, thus $\mathrm{HALT}( N_{\langle M, w\rangle}, 0)$ returns false.

Now we can see $M$ halts on $w$ if and only if $H$ accepts $\langle M, w\rangle$. Since $H$ always halts, $H$ is a decider for the normal halting problem, a contradiction!

So $\mathrm{HALT}$ does not exist.