gulp pipe(gulp.dest()) does not create any results

I had the same problem, you have to return the task inside the function:

gulp.task('default', function() {
 return gulp.src("js/*.js")
    .pipe(uglify())
    .pipe(gulp.dest('minjs'));

Also, minjs will not be a file, but a folder, were all your minified files are going to be saved.

Finally, if you want to minify only 1 file, you can specify it directly, the same with the location of the destination.

For example:

var gulp = require('gulp');
var browserify = require('browserify');
var source = require('vinyl-source-stream');
var uglify = require('gulp-uglify');    

gulp.task('browserify', function() {
    return browserify('./src/client/app.js')
        .bundle()
        //Pass desired output filename to vinyl-source-stream
        .pipe(source('main.js'))
        // Start piping stream to tasks!
        .pipe(gulp.dest('./public/'));
});

    gulp.task('build', ['browserify'], function() {
         return gulp.src("./public/main.js")
            .pipe(uglify())
            .pipe(gulp.dest('./public/'));
    });

Hope it helps!