How to check that a value has specific type

I want typescript to check value to match specific type without actually invoking this
What is the most elegant way to do this rather than I have described below?

Consider the following example:

import { OdbEventProcessorFunc } from "./OdbEventProcessor";

export function tviewEventProcessor() {
    // some implementation here... doesn't matter for the question
}

// The ugly way to do the job:
function unused_just_for_type_check_of_the_function() {
    // The following line checks that function 'tviewEventProcessor' is actually 'OdbEventProcessorFunc' and raise TS2322 error if it's not
    const unused_just_for_type_check_of_the_function2: OdbEventProcessorFunc = tviewEventProcessor; 
}

The code above does what I need, and I actually use it rare cases when I need it.
But I wonder if there is a better way to do this?

Some kind of

typescript_please_check_that(tviewEventProcessor is OdbEventProcessorFunc )

What struggles me in existing approach is:

• Its ugly and long to write
• It creates some code that can be bundled. Though it should be stripped by treeshaking, but anyway

Additional Q&A:
Q: Why doing it like this and not typecheck on caller side?
A: Because when I change definition of 'OdbEventProcessorFunc' I want IDE to navigate me with TS2322 errors not to the callers of this function, but to its definiton.

If you use an arrow function instead (which, as an expression rather than a function declaration, can be easily type-checked without any extra baggage), you can export and check the type at the same time. For example:

type Fn = (arg: string) => number;

export const fn: Fn = (arg: string) => {
    return 5;
};

If you stay with a function, you can use a satisfies that isn't used anywhere else to typecheck.

type Fn = (arg: string) => number;

export const fn = (arg: string) => {
    return 5;
};
fn satisfies Fn;

No need to wrap it in a function, or to have any emitted code there.