A variant of `Nearest`: Find the largest value that is smaller (larger) than a given value

One can use Interpolation with InterpolationOrder -> 0:

SeedRandom[1];
data = RandomReal[1, 10]
(*
  {0.817389,  0.11142,  0.789526, 0.187803, 0.241361,
   0.0657388, 0.542247, 0.231155, 0.396006, 0.700474}
*)

nf = Evaluate@ Interpolation[Transpose@{-data, data}, InterpolationOrder -> 0, 
      "ExtrapolationHandler" -> {With[{m = -Max[data]}, 
         Piecewise[{{-m, # < m}}, Indeterminate] &], 
        "WarningMessage" -> False}
      ][-#] &;

Plot[nf[x], {x, -0.5, 1.5}, 
 Epilog -> {Red, PointSize@Medium, Point[Transpose@{data, data}]}]

Replace Indeterminate with the value desired when the input falls below the minimum of the data.

Interpolation[] takes longer than Nearest[] to form the function, but it is faster to evaluate on large data:

SeedRandom[1];
data = RandomReal[1, 1000000];

nf = Evaluate@ Interpolation[Transpose@{-data, data}, InterpolationOrder -> 0, 
       "ExtrapolationHandler" -> {With[{m = -Max[data]}, 
          Piecewise[{{-m, # < m}}, Indeterminate] &], 
         "WarningMessage" -> False}
       ][-#] &; // RepeatedTiming
nf /@ RandomReal[1, 1000]; // RepeatedTiming
(*
  {1.43, Null}
  {0.0043, Null}
*)

(* Sascha's distance function  dist[]  *)
nf2 = Nearest[data, DistanceFunction -> dist]; // RepeatedTiming
nf2 /@ RandomReal[1, 2]; // RepeatedTiming
(*
  {0.000015, Null}
  {4.4, Null}
*)

Relative speed vs. length of data to evaluate the function on an input, showing that nf becomes orders of magnitude faster as the size of data increases:

Length@data   1000   10000   100000  1000000
nf2/nf         700    7000    60000   600000

The speed to form nf2 stays roughly constant. The speed to form nf is roughly linear.
The speed of nf2 seems to be improved by pre-sorting data by about 10-15%; sorting for n = 1000000 takes about 0.16 sec. on my machine.

You can use DistanceFunction as in

dist[u_, x_] := 1000000 (* some big number *) /; x > u
dist[u_, x_] := Abs[u - x] 

and use it like

Nearest[{1, 2, 2.9, 3, 4} , 2.99, DistanceFunction -> dist]
(* 2.9 *)

Unfortunately using Infinity in the first Definition of dist does yield an error

Nearest::nearuf: The user-supplied distance function dist does not give a real numeric distance when applied to the point pair 2.99` and 3.

so you have to supply an appropriately big number instead.

You can also build your own quite easily:

findL[x_, val_] := Max[Select[x, # < val &]];

This selects all the numbers in the list x less than the desired value val and then picks the largest of these. You can, of course plot:

Plot[findL[x, t], {t, 0, 1}]