# Why is O notation the worst case? [duplicate]

I don't understand why O notation is the worst case. If this notations describes a function f such that 0 <= f(n) <= cg(n), we can see that in any case f will be smaller that the original function g that describes the running time. In my idea, f(n) describes a better option because the running time is less for any input n. Someone can explain me please why O notation is the worst case?﻿

## marked as duplicate by David Richerby, Yuval Filmus algorithms StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 5 at 2:35

It doesn't, if $$f \in O(g)$$, it "just" tells us that $$g$$ is an asypmtotic upper bound for $$f$$.