gouthamphilomath wrote: $(SWE 2)$ For each $\lambda (0 < \lambda < 1$ and $\lambda = \frac{1}{n}$ for all $n = 1, 2, 3, \cdots)$, construct a continuous function $f$ such that there do not exist $x, y$ with $0 < \lambda < y = x + \lambda \le 1$ for which $f(x) = f(y).$ $f(x)=x$

I think the problem is a misstatement of part of the Universal Chord Theorem (citation from H. Boas) : A given continuous function $f$, say with domain $[0, 1]$, may have no horizontal chords at all. However, let us suppose to begin with that it does have one horizontal chord. More specifically, suppose that $f(0) = f(1)$, so that the segment $[0, 1]$ is a horizontal chord. The universal chord theorem states that there are then horizontal chords of lengths $\dfrac {1} {2}, \dfrac {1} {3}, \dfrac {1} {4}, \ldots, \dfrac {1} {n}, \ldots$, but not necessarily a horizontal chord of any given length that is not the reciprocal of an integer. I think the asked question is (not to prove the theorem!, but) to exhibit, for any $0 < \lambda < 1$ which is not the reciprocal of an integer, such an example of a continuous function having a chord of length $1$ (by $f(0) = f(1)$ for example), but does not have a chord of length $\lambda$. An example that has to be adjusted is given at Boas book

I found the description of such a function in [Hadwiger-Debrunner, Combinatorial Geometry in the Plane], attributed to P. Lévy (who also found the Universal Chord theorem). Pick $t \in (0,1)$. Define $f_t : [0,1] \to \mathbb{R}$ by $f_t(x) = \sin^2 \dfrac {\pi x} {t} - x\sin^2 \dfrac {\pi} {t}$. Since $f_t(0) = f_t(1) = 0$, the graph of the function admits a chord of length $1$ parallel to the $x$-axis. On the other hand, $f_t(x+t) = f_t(x) - t\sin^2 \dfrac {\pi} {t}$, so it follows that $f_t(x+t) \neq f_t(x)$ when $t \not \in \left \{ 1, \dfrac {1} {2}, \dfrac {1} {3}, \ldots \right \}$, hence in this case the graph of the function admits no chord of length $t$ parallel to the $x$-axis.